Q.At what point is the slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ maximum? Also find the maximum slope.

Showing the 12 most recent of 22 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If a straight line $y = mx + c$ touches the circle $x^2 + y^2 = 4$ and the parabola $y^2 = 4x$, then $2m^2 =$ (A) $\sqrt{2} + 1$ (B) $2$ (C) $\dfrac{1}{2}$ (D) $\sqrt{2} - 1$

   ›Reveal solutionSolution

   Combine the tangency conditions for the circle and parabola: $4m^4 + 4m^2 - 1 = 0$ gives $2m^2 = \sqrt{2} - 1$.

   Let the line be $y = mx + c$.

   Tangent to the parabola $y^2 = 4x$ (here $a = 1$): the condition is

   $c = \frac{a}{m} = \frac{1}{m}.$

   Tangent to the circle $x^2 + y^2 = 4$ (radius $2$): the distance from the origin equals the radius,

   $\frac{|c|}{\sqrt{1 + m^2}} = 2 \ \Rightarrow\ c^2 = 4(1 + m^2).$

   Substitute $c = \dfrac{1}{m}$:

   $\frac{1}{m^2} = 4(1 + m^2) \ \Rightarrow\ 1 = 4m^2 + 4m^4.$

   Let $u = m^2$: …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.$(1,1)$ is the focus of the parabola $y^2 - 4ax - 2ay + a^2 = 0$. If the circles $(x-\alpha)^2 + (y-\beta)^2 = r^2$ ($\alpha, \beta$ are parameters) touch the X-axis and the axis of the given parabola, then $\{(\alpha, \beta) | \alpha, \beta \in \mathbb{R}\}$ is (A) a line $y = \dfrac{1}{2}$ (B) a line $y = 1$ (C) a circle $x^2 + y^2 = \dfrac{1}{4}$ (D) a parabola $y^2 = 2x$

   ›Reveal solutionSolution

   The given parabola’s axis is vertical, so the family of circles touching both the X‑axis and that axis must have centres lying on the line halfway between them — which turns out to be $y = \tfrac12$. The correct option is (A).

   ---

   We start with the parabola

   $y^2 - 4ax - 2ay + a^2 = 0.$

   The focus is given as $(1,1)$. That will let us find $a$, and then the axis of the parabola. The circles in question touch the X‑axis and the axis of the parabola; their centres $(\alpha,\beta)$ must satisfy a simple geometric condition.

   ---

   1. Find $a$ from the focus condition

   Rewrite the parabola by completing the square in $y$:

   $y^2 - 2ay + a^2 - 4ax = 0 \quad\Longrightarrow\quad (y - a)^2 = 4ax.$

   So the equation is

   $(y - a)^2 = 4a x.$

   This is a standard right‑opening parabola with vertex at $(0,a)$ and focus at $(a, a)$.

   We are told the focus is $(1,1)$, so

   $a = 1.$

   Thus the parabola is

   $(y - 1)^2 = 4x,$

   with vertex $(0,1)$, focus $(1,1)$, and axis the horizontal line $y = 1$.

   Watch out

   A common mistake: thinking the axis is vertical because the $y$‑term is squared. Actually $(y-1)^2 = 4x$ opens to the right, so its axis is horizontal: $y = 1$.

   ---

   2. Interpret the circle condition

   We have circles

   $(x - \alpha)^2 + (y - \beta)^2 = r^2$

   that touch (are tangent to) two lines:

   • The X‑axis: $y = 0$.
   • The axis of the parabola: $y = 1$.

   For a circle to be tangent to a horizontal line $y = c$, the vertical distance from the centre $(\alpha,\beta)$ to that line must equal the radius $r$.

   • Tangent to $y=0$: $\beta - 0 = r$ (since $\beta > 0$ for the circle to lie above the X‑axis).
   • Tangent to $y=1$: $| \beta - 1 | = r$.

   Since the circle touches both lines, both distances equal the same $r$.

   ---

   3. Equate the distances

   From the two conditions:

   $\beta = r \quad\text{and}\quad |\beta - 1| = r.$

   Thus

   $$$$

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3. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the roots of the equation $32x^3 - 48x^2 + 22x - 3 = 0$ are in arithmetic progression, then the square of the common difference of the roots is (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{16}$ (C) $\dfrac{1}{9}$ (D) $\dfrac{1}{25}$

   ›Reveal solutionSolution

   For a cubic with roots in arithmetic progression, the middle root is the average of the three, which equals one-third the sum of the roots. Using Vieta’s formulas, we find the middle root, then the common difference, and square it to get $\frac{1}{16}$.

   We are told the roots of $32x^3 - 48x^2 + 22x - 3 = 0$ are in arithmetic progression. That means they can be written as $a-d$, $a$, $a+d$, where $a$ is the middle term and $d$ is the common difference. The key insight: when roots are equally spaced, the middle root is simply the average of the three roots, which is also one-third of their sum. Vieta’s formulas give us that sum directly from the coefficients, so we can find $a$ immediately. Then we can use another Vieta relation to solve for $d^2$.

   1. Write the roots in AP form

      Let the roots be $p-d$, $p$, $p+d$. Their sum is $(p-d) + p + (p+d) = 3p$.

   2. Use Vieta for the sum of roots

      For $32x^3 - 48x^2 + 22x - 3 = 0$, the sum of roots (with sign) is $-\frac{\text{coefficient of }x^2}{\text{coefficient of }x^3} = -\frac{-48}{32} = \frac{48}{32} = \frac{3}{2}$.

      So $3p = \frac{3}{2}$, giving $p = \frac{1}{2}$.

   3. Use Vieta for the sum of pairwise products

      The sum of products taken two at a time is $\frac{\text{coefficient of }x}{\text{coefficient of }x^3} = \frac{22}{32} = \frac{11}{16}$.

      In terms of $p$ and $d$:

      $(p-d)p + p(p+d) + (p-d)(p+d) = p(p-d) + p(p+d) + (p^2 - d^2)$.

      Simplify:

      $p^2 - pd + p^2 + pd + p^2 - d^2 = 3p^2 - d^2$.

      So $3p^2 - d^2 = \frac{11}{16}$.

   4. Substitute $p = \frac{1}{2}$

      $3\left(\frac{1}{2}\right)^2 - d^2 = \frac{11}{16}$

      $\Rightarrow 3 \cdot \frac{1}{4} - d^2 = \frac{11}{16}$ …

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4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the sum of two roots of the equation $x^4 - 2x^3 + x^2 + 4x - 6 = 0$ is zero then the sum of the squares of the other two roots is (A) $-6$ (B) $1$ (C) $-2$ (D) $0$

   ›Reveal solutionSolution

   Using the condition that two roots sum to zero, we factor the quartic into quadratics, find the remaining roots, and compute the sum of squares of the other two roots as $1$.

   We are given the quartic equation

   $x^4 - 2x^3 + x^2 + 4x - 6 = 0$

   and told that the sum of two of its roots is zero. That means if those two roots are $a$ and $b$, then $a + b = 0$, so $b = -a$. The problem asks for the sum of the squares of the other two roots.

   Concept & Intuition

   When a polynomial has a known relationship among its roots, we can often factor it into smaller-degree polynomials whose coefficients are determined by that relationship. Here, if two roots are opposites, then the quartic must be divisible by a quadratic factor of the form $x^2 - r^2$ (since $(x - a)(x + a) = x^2 - a^2$). The remaining quadratic factor will give the other two roots, and we can find their sum of squares directly from its coefficients — no need to solve for individual roots explicitly.

   ---

   Step-by-step solution

   1. Set up the factorization Let the four roots be $a, -a, p, q$. Then the quartic can be written as

   $(x - a)(x + a)(x - p)(x - q) = (x^2 - a^2)(x^2 - (p+q)x + pq).$

   Our goal is to find $p^2 + q^2$.

   2. Expand the product Multiply out:

   $(x^2 - a^2)(x^2 - Sx + P) \quad \text{where } S = p+q,\; P = pq.$

   This gives

   $x^4 - Sx^3 + Px^2 - a^2 x^2 + a^2 S x - a^2 P.$

   Collect like terms:

   $x^4 - Sx^3 + (P - a^2)x^2 + (a^2 S)x - a^2 P.$

   3. Match coefficients with the given polynomial The given polynomial is

   $x^4 - 2x^3 + x^2 + 4x - 6.$

   Equating coefficients:

   • Coefficient of $x^3$: $-S = -2 \implies S = 2$.
   • Coefficient of $x^2$: $P - a^2 = 1$. …

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5. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If $m_1$ and $m_2$ are the slopes of the tangents drawn from the point $(1,4)$ to the parabola $y^2 = 11x$ then $2(m_1^2 + m_2^2) =$ (A) $18$ (B) $21$ (C) $24$ (D) $22$

   ›Reveal solutionSolution

   The key idea is to use the equation of a tangent to the parabola $y^2 = 11x$ in slope form, impose that it passes through $(1,4)$, and then use the quadratic in $m$ to find $m_1^2 + m_2^2$ without solving individually. The final result is $2(m_1^2 + m_2^2) = 22$.

   ---

   We are given the parabola $y^2 = 11x$. For a parabola of the form $y^2 = 4ax$, the slope $m$ of a tangent that touches it satisfies the equation $y = mx + \frac{a}{m}$. Here $4a = 11$, so $a = \frac{11}{4}$. Thus any tangent (with slope $m \neq 0$) to this parabola has equation:

   $y = mx + \frac{11}{4m}.$

   We want the tangents that pass through the external point $(1,4)$. Substituting $x=1$, $y=4$ gives:

   $4 = m(1) + \frac{11}{4m}.$

   Multiply through by $4m$ (valid since $m \neq 0$):

   $16m = 4m^2 + 11.$

   Rearrange into a standard quadratic:

   $4m^2 - 16m + 11 = 0.$

   This quadratic has two roots $m_1$ and $m_2$, which are the slopes of the two tangents from $(1,4)$ to the parabola.

   Now we need $2(m_1^2 + m_2^2)$. Instead of solving for $m_1$ and $m_2$ individually, we use the relations from the quadratic:

   $m_1 + m_2 = \frac{16}{4} = 4, \quad m_1 m_2 = \frac{11}{4}.$

   Recall the identity:

   $m_1^2 + m_2^2 = (m_1 + m_2)^2 - 2m_1 m_2.$

   Substitute:

   $$$$

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6. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.The numerically greatest term in the expansion of $(3x-16y)^{15}$ when $x = \dfrac{2}{3}$ and $y = \dfrac{3}{2}$ is (A) 13th term (B) 14th term (C) 15th term (D) 16th term

   ›Reveal solutionSolution

   The numerically greatest term in the binomial expansion is found by comparing successive term ratios; for the given substitution, the 14th term is the largest, so the answer is option (B).

   We are asked for the numerically greatest term in the expansion of $(3x - 16y)^{15}$ when $x = \frac{2}{3}$ and $y = \frac{3}{2}$.

   The key idea: In a binomial expansion $(a + b)^n$, the terms increase in magnitude up to a point and then decrease. By examining the ratio of consecutive terms, we can locate where the maximum occurs — without computing all 16 terms.

   ---

   1. Substitute the given values and simplify the expression

   First, plug in $x = \frac{2}{3}$ and $y = \frac{3}{2}$:

   $3x = 3 \cdot \frac{2}{3} = 2, \quad 16y = 16 \cdot \frac{3}{2} = 24$

   So the expression becomes:

   $(3x - 16y)^{15} = (2 - 24)^{15} = (-22)^{15}$

   That’s just a single number — but wait: the expansion’s terms are not all equal; they come from the binomial expansion of $(3x - 16y)^{15}$ before substitution. We must substitute into the general term.

   ---

   2. Write the general term

   The general term in $(A + B)^{15}$ is:

   $T_{r+1} = \binom{15}{r} A^{15-r} B^r$

   Here $A = 3x$ and $B = -16y$. So:

   $T_{r+1} = \binom{15}{r} (3x)^{15-r} (-16y)^r$

   Substitute $x = \frac{2}{3}$, $y = \frac{3}{2}$:

   $T_{r+1} = \binom{15}{r} (2)^{15-r} (-24)^r$

   So the magnitude (absolute value) is:

   $|T_{r+1}| = \binom{15}{r} \cdot 2^{15-r} \cdot 24^r$

   ---

   3. Find the ratio of successive terms

   Let $t_r = |T_{r+1}|$. Then:

   $\frac{t_{r+1}}{t_r} = \frac{\binom{15}{r+1} 2^{14-r} 24^{r+1}}{\binom{15}{r} 2^{15-r} 24^r}$

   Simplify:

   $\frac{\binom{15}{r+1}}{\binom{15}{r}} = \frac{15 - r}{r+1}, \quad \frac{2^{14-r}}{2^{15-r}} = \frac{1}{2}, \quad \frac{24^{r+1}}{24^r} = 24$

   Thus:

   $\frac{t_{r+1}}{t_r} = \frac{15 - r}{r+1} \cdot \frac{24}{2} = \frac{15 - r}{r+1} \cdot 12$

   ---

   4. Determine when terms increase or decrease

   Terms increase as long as $\frac{t_{r+1}}{t_r} > 1$:

   $\frac{15 - r}{r+1} \cdot 12 > 1 \quad \Rightarrow \quad 12(15 - r) > r + 1$

   $180 - 12r > r + 1 \quad \Rightarrow \quad 179 > 13r \quad \Rightarrow \quad r < \frac{179}{13} \approx 13.769$

   So for $r = 0, 1, \dots, 13$, the ratio is > 1, meaning terms increase up to $r = 13$.

   At $r = 13$: $\frac{t_{14}}{t_{13}} = \frac{15-13}{13+1} \cdot 12 = \frac{2}{14} \cdot 12 = \frac{24}{14} \approx 1.714 > 1$, so $t_{14} > t_{13}$.

   At $r = 14$: $\frac{t_{15}}{t_{14}} = \frac{15-14}{14+1} \cdot 12 = \frac{1}{15} \cdot 12 = 0.8 < 1$, so $t_{15} < t_{14}$. …

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7. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.The maximum value of the function $f(x) = 3\sin^{12}x + 4\cos^{16}x$ is (A) 4 (B) 5 (C) 6 (D) 7

   ›Reveal solutionSolution

   The key idea is to bound each trigonometric term by its maximum possible value (1 for sine or cosine squared) and then check if both maxima can occur simultaneously. The maximum value is 4, corresponding to option (A).

   We want the maximum of $f(x) = 3\sin^{12}x + 4\cos^{16}x$. Since $\sin^2 x$ and $\cos^2 x$ are always between 0 and 1, raising them to higher powers only makes them smaller or keeps them the same. So the largest each term can be is when the base is 1.

   Intuition:

   If $\sin^2 x = 1$, then $\sin^{12}x = 1$ and $\cos^2 x = 0$, so $\cos^{16}x = 0$. That gives $f = 3$.

   If $\cos^2 x = 1$, then $\cos^{16}x = 1$ and $\sin^2 x = 0$, giving $f = 4$.

   So 4 is already larger than 3. Could we get more than 4? For that, both $\sin^{12}x$ and $\cos^{16}x$ would need to be positive simultaneously, but then each is less than 1, so the weighted sum might exceed 4? Let’s check carefully.

   Step-by-step reasoning:

   1. Bound each term individually For any real $x$, $0 \le \sin^2 x \le 1$ and $0 \le \cos^2 x \le 1$. Since $12$ and $16$ are even positive integers,

   $0 \le \sin^{12}x \le 1, \quad 0 \le \cos^{16}x \le 1.$

   Hence

   $f(x) = 3\sin^{12}x + 4\cos^{16}x \le 3\cdot 1 + 4\cdot 1 = 7.$

   But this bound is not attainable because $\sin^{12}x$ and $\cos^{16}x$ cannot both be 1 at the same time (since $\sin^2 x + \cos^2 x = 1$).

   2. Find when each term individually reaches its maximum

      • $\sin^{12}x = 1$ when $\sin^2 x = 1$, i.e., $x = \frac{\pi}{2} + k\pi$. Then $\cos^2 x = 0$, so $\cos^{16}x = 0$. At such $x$, $f = 3\cdot 1 + 4\cdot 0 = 3$.
      • $\cos^{16}x = 1$ when $\cos^2 x = 1$, i.e., $x = k\pi$. Then $\sin^2 x = 0$, so $\sin^{12}x = 0$. At such $x$, $f = 3\cdot 0 + 4\cdot 1 = 4$.

      So far, the largest value we have is 4.

   3. Could a mix give more than 4?

      Suppose both $\sin^2 x$ and $\cos^2 x$ are positive. Let $a = \sin^2 x$, $b = \cos^2 x$, with $a + b = 1$, $a, b \ge 0$.

      Then

      $f = 3a^6 + 4b^8.$ …

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8. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.The equation of the common tangent to the parabola $y^2 = 8x$ and the circle $x^2 + y^2 = 2$ is $ax + by + 2 = 0$. If $-\dfrac{a}{b} > 0$, then $3a^2 + 2b + 1 =$ (A) $5$ (B) $4$ (C) $3$ (D) $2$

   ›Reveal solutionSolution

   The common tangent to the parabola $y^2 = 8x$ and the circle $x^2 + y^2 = 2$ has the form $y = mx + \frac{2}{m}$ (for the parabola) and must satisfy the circle’s tangency condition, leading to $m = 1$ (since $-\frac{a}{b} > 0$ forces the positive slope). This gives $a = 1, b = -1$, so $3a^2 + 2b + 1 = 2$.

   ---

   Concept & Intuition

   When two curves share a common tangent, the line must satisfy the tangency condition for both curves simultaneously. For a parabola $y^2 = 4ax$, the family of tangents in slope form is $y = mx + \frac{a}{m}$. For a circle, the condition that a line $y = mx + c$ is tangent is that the perpendicular distance from the center equals the radius. Matching these gives the slope, and the sign condition picks the correct one.

   ---

   Step-by-step solution

   1. Identify the parabola’s tangent family The parabola is $y^2 = 8x$, so $4a = 8 \Rightarrow a = 2$. Any tangent to this parabola (with slope $m \neq 0$) is

   $y = mx + \frac{2}{m}.$

   This is the standard formula $y = mx + \frac{a}{m}$ for $y^2 = 4ax$.

   2. Apply the circle’s tangency condition The circle is $x^2 + y^2 = 2$, center $(0,0)$, radius $r = \sqrt{2}$. For the line $y = mx + \frac{2}{m}$ to be tangent to the circle, the distance from the center to the line must equal $\sqrt{2}$. Rewrite the line as:

   $mx - y + \frac{2}{m} = 0.$

   Distance from $(0,0)$ is

   $\frac{\left| \frac{2}{m} \right|}{\sqrt{m^2 + 1}} = \sqrt{2}.$

   3. Solve for $m$ Square both sides:

   $\frac{4}{m^2 (m^2 + 1)} = 2 \quad\Rightarrow\quad \frac{4}{m^2(m^2+1)} = 2.$

   Multiply: $4 = 2m^2(m^2+1)$ $\Rightarrow$ $2 = m^2(m^2+1)$.

   Let $t = m^2$:

   $t(t+1) = 2 \quad\Rightarrow\quad t^2 + t - 2 = 0 \quad\Rightarrow\quad (t+2)(t-1)=0.$

   So $t = 1$ or $t = -2$ (reject negative). Hence $m^2 = 1$, so $m = \pm 1$.

   4. Use the sign condition $-\frac{a}{b} > 0$ The given tangent is $ax + by + 2 = 0$. Compare with $y = mx + \frac{2}{m}$. Rewrite $y = mx + \frac{2}{m}$ as $mx - y + \frac{2}{m} = 0$. …

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9. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If $f(x) = (2x-1)(3x+2)(4x-3)$ is a real valued function defined on $\left[\frac{1}{2}, \frac{3}{4}\right]$, then the value(s) of ‘c’ as defined in the statement of Rolle’s theorem (A) Does not exist (B) $\dfrac{7 \pm \sqrt{247}}{36}$ (C) $\dfrac{7 - \sqrt{247}}{36}$ (D) $\dfrac{7 + \sqrt{247}}{36}$

   ›Reveal solutionSolution

   Rolle's theorem applies since $f\!\left(\tfrac12\right)=f\!\left(\tfrac34\right)=0$; solving $f'(c)=72c^2-28c-11=0$ gives $c=\dfrac{7\pm\sqrt{247}}{36}$, and only $\dfrac{7+\sqrt{247}}{36}\approx0.63$ lies in $\left(\tfrac12,\tfrac34\right)$. Answer: (D).

   Concept

   Rolle's theorem: if $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then some $c\in(a,b)$ has $f'(c)=0$. The valid $c$ must lie strictly inside the interval.

   Solution

   1. Check the hypotheses. $f(x)=(2x-1)(3x+2)(4x-3)$ is a polynomial, hence continuous and differentiable everywhere. The endpoints are zeros of two factors:

   $f\!\left(\tfrac12\right)=0\ \text{(from }2x-1),\qquad f\!\left(\tfrac34\right)=0\ \text{(from }4x-3),$

   so $f\!\left(\tfrac12\right)=f\!\left(\tfrac34\right)$ and Rolle's theorem applies.

   2. Derivative (product rule on three factors, $u'=2,\ v'=3,\ w'=4$):

   $f'(x)=2(3x+2)(4x-3)+3(2x-1)(4x-3)+4(2x-1)(3x+2).$

   Expanding and adding, each term contributes $24x^2$, so

   $f'(x)=72x^2-28x-11.$

   3. Solve $f'(c)=0$. …

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10. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.If the extremities of the latus recta having positive ordinate of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $(a > b)$ lie on the parabola $x^2 + 2ay - 4 = 0$, then the points $(a, b)$ lie on the curve (A) $xy = 4$ (B) $x^2 + y^2 = 4$ (C) $\dfrac{x^2}{4} + \dfrac{y^2}{1} = 1$ (D) $\dfrac{x^2}{1} + \dfrac{y^2}{4} = 1$

    ›Reveal solutionSolution

    The extremities of the latus recta with positive y-coordinate of the ellipse are points where the ellipse’s focal chord meets the ellipse; substituting these into the parabola’s equation gives a relation between $a$ and $b$, which simplifies to $a^2 + b^2 = 4$, so $(a,b)$ lies on the circle $x^2 + y^2 = 4$.

    Concept & Intuition

    The problem connects two conics: an ellipse and a parabola. The “extremities of the latus recta having positive ordinate” means the upper endpoints of the two latus recta (the chords through each focus perpendicular to the major axis). For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b$, the foci are at $(\pm ae,0)$ where $e=\sqrt{1-\frac{b^2}{a^2}}$. The latus rectum through the right focus has endpoints $\left(ae, \pm \frac{b^2}{a}\right)$; the one through the left focus has $\left(-ae, \pm \frac{b^2}{a}\right)$. The “positive ordinate” picks the upper halves: $(ae, b^2/a)$ and $(-ae, b^2/a)$. These two points must satisfy the parabola $x^2+2ay-4=0$. Substituting gives two equations that must both hold, leading to a relation between $a$ and $b$.

    ---

    1. Identify the points For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b$, the eccentricity is $e=\sqrt{1-\frac{b^2}{a^2}}$. The foci are $(\pm ae,0)$. The length of the latus rectum is $\frac{2b^2}{a}$, so the endpoints of the latus rectum through the right focus are $\left(ae, \pm \frac{b^2}{a}\right)$. Taking the positive ordinate gives the point

    $P = \left(ae, \frac{b^2}{a}\right).$

    Similarly, for the left focus, the upper endpoint is

    $Q = \left(-ae, \frac{b^2}{a}\right).$

    2. Apply the parabola condition The parabola is $x^2 + 2ay - 4 = 0$. Both $P$ and $Q$ must lie on it. Substituting $P$:

    $(ae)^2 + 2a\left(\frac{b^2}{a}\right) - 4 = 0 \quad\Rightarrow\quad a^2 e^2 + 2b^2 - 4 = 0.$

    Substituting $Q$ gives the same equation because $x^2$ is the same:

    $(-ae)^2 + 2a\left(\frac{b^2}{a}\right) - 4 = a^2 e^2 + 2b^2 - 4 = 0.$

    So we have one condition:

    $a^2 e^2 + 2b^2 = 4.$

    3. Replace $e^2$ using the ellipse relation …

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11. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.$(1,1)$ is the vertex and $x+y+1=0$ is the directrix of a parabola. If $(a,b)$ is its focus and $(c,d)$ is the point of intersection of the directrix and the axis of the parabola, then $a+b+c+d=$ (A) $6$ (B) $5$ (C) $4$ (D) $3$

    ›Reveal solutionSolution

    The vertex and directrix fix the parabola’s axis and focus; using the vertex as the midpoint of the focus and the directrix’s foot on the axis gives the focus and foot coordinates, whose sum is 4.

    Concept & Intuition

    For any parabola, the vertex lies exactly halfway between the focus and the directrix along the axis. That means the vertex is the midpoint of the segment joining the focus to the point where the axis meets the directrix. If we know the vertex and the directrix line, we can find the axis (perpendicular to the directrix through the vertex), then find the foot of the vertex on the directrix, and finally use the midpoint property to locate the focus. Adding the coordinates of the focus and that foot gives the required sum.

    ---

    1. Find the axis of the parabola. The axis is the line through the vertex $(1,1)$ perpendicular to the directrix $x + y + 1 = 0$. The directrix has slope $-1$, so the axis has slope $1$. Equation of the axis:

    $y - 1 = 1(x - 1) \quad\Rightarrow\quad y = x.$

    2. Find the foot of the vertex on the directrix (point $(c,d)$). This is the intersection of the axis $y = x$ with the directrix $x + y + 1 = 0$. Substitute $y = x$:

    $x + x + 1 = 0 \quad\Rightarrow\quad 2x = -1 \quad\Rightarrow\quad x = -\frac12.$

    Then $y = -\frac12$. So

    $(c,d) = \left(-\frac12, -\frac12\right).$

    3. Use the vertex as the midpoint of the focus and this foot. Let the focus be $(a,b)$. The vertex $(1,1)$ is the midpoint of $(a,b)$ and $\left(-\frac12, -\frac12\right)$:

    $\frac{a + (-\frac12)}{2} = 1, \qquad \frac{b + (-\frac12)}{2} = 1.$

    Solve:

    $a - \frac12 = 2 \quad\Rightarrow\quad a = \frac52,$ …

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12. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.If the focal distance of a point $P(2, y_1)$ on the parabola $y^2 = kx$ is $3$, then the equation of the tangent drawn at $P$ to the given parabola is (A) $x \pm 2\sqrt{2}y + 4 = 0$ (B) $x \pm 2\sqrt{2}y + 2 = 0$ (C) $x \pm \sqrt{2}y + 4 = 0$ (D) $x \pm \sqrt{2}y + 2 = 0$

    ›Reveal solutionSolution

    The focal distance condition gives the point’s coordinates; substituting into the parabola yields the parameter, then the tangent equation is derived and matched to the given options. The correct option is (A).

    Concept & Intuition

    For a parabola $y^2 = kx$, the focus is at $(\frac{k}{4}, 0)$. The focal distance of a point is its distance from the focus. Knowing this distance lets us find the unknown coordinate $y_1$ and the constant $k$. Once we have the point $P$ and the parabola, we can write the tangent equation using the standard formula for a tangent to $y^2 = 4ax$. The multiple-choice options suggest the tangent lines come in a pair (symmetric about the x‑axis), which matches the symmetry of the parabola.

    Step‑by‑step solution

    1. Identify the parabola’s standard form

       The given parabola is $y^2 = kx$. Compare with the standard form $y^2 = 4ax$.

       Hence $4a = k$ so $a = \frac{k}{4}$.

       The focus is at $(a, 0) = \left(\frac{k}{4}, 0\right)$.

    2. Use the focal distance condition

       Point $P$ is $(2, y_1)$. Its distance to the focus is given as 3:

    $\sqrt{\left(2 - \frac{k}{4}\right)^2 + (y_1 - 0)^2} = 3.$

    Square both sides:

    $\left(2 - \frac{k}{4}\right)^2 + y_1^2 = 9. \qquad(1)$

    3. Use that $P$ lies on the parabola Since $P(2, y_1)$ satisfies $y^2 = kx$, we have

    $y_1^2 = k \cdot 2 = 2k. \qquad(2)$

    4. Solve for $k$ Substitute (2) into (1):

    $\left(2 - \frac{k}{4}\right)^2 + 2k = 9.$

    Expand:

    $4 - k + \frac{k^2}{16} + 2k = 9 \quad\Rightarrow\quad 4 + k + \frac{k^2}{16} = 9.$

    Multiply by 16:

    $64 + 16k + k^2 = 144 \quad\Rightarrow\quad k^2 + 16k - 80 = 0.$

    Solve:

    $k = \frac{-16 \pm \sqrt{256 + 320}}{2} = \frac{-16 \pm \sqrt{576}}{2} = \frac{-16 \pm 24}{2}.$

    So $k = 4$ or $k = -20$.

    Since the focal distance is positive and the parabola opens to the right for positive $k$ (and the point has $x=2>0$), we take $k = 4$.

    (If $k = -20$, the parabola opens left and the focus would be at $(-5,0)$; the distance condition could still hold, but the tangent options given are symmetric and positive‑looking, so $k=4$ is the intended case.)

    5. Find $y_1$

       From (2): $y_1^2 = 2 \cdot 4 = 8$ so $y_1 = \pm 2\sqrt{2}$.

       Thus $P$ is $(2, 2\sqrt{2})$ or $(2, -2\sqrt{2})$.

    6. Write the tangent equation …

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