(a) Show that the equation of the parabola with focus $(-\sqrt2, 0)$ and directrix $x=\sqrt2$ is $y^2=-4\sqrt2 x$. **OR** (b) Find the value of $\cot^{-1}(1)+\sin^{-1}\left(-\dfrac{\sqrt3}{2}\right)-\sec^{-1}(-\sqrt2)$.

> [!TLDR] > (a) derives the Cartesian equation of a left-opening parabola from the focus-directrix definition; (b) evaluates a combination of inverse trigonometric functions using their principal-value ranges. > > **(a) Equation of the parabola** > 1. For a point $P(x,y)$ on the parabola, distance to focus $S(-\sqrt2,0)$ equals distance to directrix $x=\sqrt2$. > 2. $\sqrt{(x+\sqrt2)^2+y^2}=|x-\sqrt2|$. > 3. Square both sides: $(x+\sqrt2)^2+y^2=(x-\sqrt2)^2$. > 4. Expand: $x^2+2\sqrt2x+2+y^2=x^2-2\sqrt2x+2$. > 5. Cancel $x^2$ and $2$ from both sides: $y^2=-4\sqrt2x$, the required equation. > > **(b) Evaluating cot^{-1}(1) + sin^{-1}(-root3/2) - sec^{-1}(-root2)** > 1. $\cot^{-1}(1)$: principal range $(0,\pi)$; $\cot\frac\pi4=1 \Rightarrow \cot^{-1}(1)=\frac\pi4$. > 2. $\sin^{-1}\!\left(-\frac{\sqrt3}2\right)$: principal range $\left[-\frac\pi2,\frac\pi2\right]$; $\sin\left(-\frac\pi3\right)=-\frac{\sqrt3}2 \Rightarrow$ value $=-\frac\pi3$. …

(a) Show that the equation of the parabola with focus (-√2, 0) and directrix x=√2 is y²=-4√2 x. **OR** (b) Fin

Question 109 of 126

(-\sqrt2, 0)

(b) Find the value of

\cot^{-1}(1)+\sin^{-1}\left(-\dfrac{\sqrt3}{2}\right)-\sec^{-1}(-\sqrt2)

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2022Subjective· 5mImportance★★★★★

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Concept understanding — Parabola — Standard Equation

Setting eccentricity $e=1$ in the conic definition ( $SP=e\cdot PM$ ) gives the parabola — the locus of points equidistant from a fixed focus and a fixed directrix.

Standard form (vertex at the origin, focus on the axis). With focus $(a,0)$ and directrix $x=-a$ : $y^2=4ax$ (opens right). The other three orientations, by symmetry: $y^2=-4ax$ (left), $x^2=4ay$ (up), $x^2=-4ay$ (down).

Vocabulary. Axis: line through the focus $\perp$ directrix. Vertex: where the axis meets the curve. Focal chord: any chord through the focus. Latus rectum: the focal chord $\perp$ the axis; for $y^2=4ax$ its endpoints are $(a,\pm2a)$ , so its length is $4a$ .

Vertex at $(h,k)$ :

Equation	Focus	Directrix
$(y-k)^2=4a(x-h)$	$(h+a,k)$	$x=h-a$
$(y-k)^2=-4a(x-h)$	$(h-a,k)$	$x=h+a$
$(x-h)^2=4a(y-k)$	$(h,k+a)$	$y=k-a$
$(x-h)^2=-4a(y-k)$	$(h,k-a)$	$y=k+a$

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(a) derives the Cartesian equation of a left-opening parabola from the focus-directrix definition; (b) evaluates a combination of inverse trigonometric functions using their principal-value ranges.

(a) Equation of the parabola

For a point $P(x,y)$ on the parabola, distance to focus $S(-\sqrt2,0)$ equals distance to directrix $x=\sqrt2$ .
$\sqrt{(x+\sqrt2)^2+y^2}=|x-\sqrt2|$ .
Square both sides: $(x+\sqrt2)^2+y^2=(x-\sqrt2)^2$ .
Expand: $x^2+2\sqrt2x+2+y^2=x^2-2\sqrt2x+2$ .
Cancel $x^2$ and $2$ from both sides: $y^2=-4\sqrt2x$ , the required equation.

(b) Evaluating cot^{-1}(1) + sin^{-1}(-root3/2) - sec^{-1}(-root2)

$\cot^{-1}(1)$ : principal range $(0,\pi)$ ; $\cot\frac\pi4=1 \Rightarrow \cot^{-1}(1)=\frac\pi4$ .
$\sin^{-1}\!\left(-\frac{\sqrt3}2\right)$ : principal range $\left[-\frac\pi2,\frac\pi2\right]$ ; $\sin\left(-\frac\pi3\right)=-\frac{\sqrt3}2 \Rightarrow$ value $=-\frac\pi3$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2022Set ANNUAL1 markMCQ
$x^2=24y$
(a) $8$
(b) $24$
(c) $6$
(d) $12$
›Reveal solutionSolution
Comparing $x^2=24y$ with the standard form $x^2=4ay$ directly gives the latus rectum length as $24$ .

The standard form of a parabola opening upward is $x^2=4ay$ , whose latus rectum has length $4a$ .

Comparing the given equation $x^2=24y$ with $x^2=4ay$ , we get $4a=24$ . …
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CBSE 2020Set ANNUAL1 markMCQ
$x^2 = 8y - 1$
(a) $\left(0, -\dfrac{1}{8}\right)$
(b) $\left(-\dfrac{1}{8}, 0\right)$
(c) $\left(\dfrac{1}{8}, 0\right)$
(d) $\left(0, \dfrac{1}{8}\right)$
›Reveal solutionSolution
Rewriting $x^2=8y-1$ in the standard form $x^2=4a(y-k)$ shows the vertex is $\left(0,\dfrac{1}{8}\right)$ .

Start with $x^2=8y-1$ .

Add $1$ to both sides and factor the right side: $x^2 = 8y-1 = 8\left(y-\dfrac{1}{8}\right)$ .

This is now in the standard vertical-axis parabola form $x^2=4a(y-k)$ , where the vertex is $(h,k)=(0,k)$ (here $h=0$ since there is no $x$ -shift, only a coefficient of $x^2$ ). …
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CBSE 2016Set ANNUAL1 markMCQ
$(2,-3)$
(a) $2$
(b) $4$
(c) $6$
(d) $8$
›Reveal solutionSolution
The length of the latus rectum is $4a=8$ , where $a$ is the vertex-to-directrix distance.

For a parabola, the vertex lies midway between the focus and the directrix, at distance $a$ from each.

Vertex $=(2,-3)$ , directrix $x=4$ (a vertical line), so the axis of the parabola is horizontal, $y=-3$ .

Distance from vertex to directrix: $a=|4-2|=2$ .

Length of latus rectum $=4a=4(2)=8$ . …
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Q.(a) Show that the equation of the parabola with focus (−2,0)(-\sqrt2, 0)(−2​,0) and directrix x=2x=\sqrt2x=2​ is y2=−42xy^2=-4\sqrt2 xy2=−42​x. OR

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Q.(a) Show that the equation of the parabola with focus $(-\sqrt2, 0)$ and directrix $x=\sqrt2$ is $y^2=-4\sqrt2 x$ . OR

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