Q.Show that the line $x-y+4=0$ is a tangent to the ellipse $x^2+3y^2=12$. Also find the coordinates of the point of contact.

The condition for a line $y=mx-2am-am^3$ to be a normal through a given external point is cubic in $m$, giving 3 normals.

1. The normal to $y^2=4ax$ at the parametric point $(at^2,2at)$ has equation $y=-tx+2at+at^3$, i.e. in terms of slope $m=-t$: $y=mx-2am-am^3$.
2. For this normal to pass through a fixed external point $(x_1,y_1)$: $y_1=mx_1-2am-am^3$, i.e. $am^3+(2a-x_1)m+y_1=0$.
3. This is a cubic equation in $m$, so it has (up to) $3$ real roots, each giving a distinct normal from the point.
4. Hence, in general, exactly $3$ normals can be drawn from a point to the parabola $y^2=4ax$.

Using the tangent condition for a hyperbola, $k=\dfrac{9}{4}$.

1. The hyperbola $4x^2-y^2=36$ can be written as $\dfrac{x^2}{9}-\dfrac{y^2}{36}=1$, giving $a^2=9$ and $b^2=36$.
2. Rewrite the line $5x-2y+4k=0$ in slope form: $2y=5x+4k \Rightarrow y=\dfrac{5}{2}x+2k$, so $m=\dfrac{5}{2}$ and $c=2k$.
3. The condition for the line $y=mx+c$ to be a tangent to $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is $c^2=a^2m^2-b^2$.
4. Substituting: $(2k)^2=9\left(\dfrac{5}{2}\right)^2-36=9\cdot\dfrac{25}{4}-36=\dfrac{225}{4}-\dfrac{144}{4}=\dfrac{81}{4}$.
5. So $4k^2=\dfrac{81}{4}\Rightarrow k^2=\dfrac{81}{16}\Rightarrow k=\pm\dfrac{9}{4}$.
6. Taking the value matching the given options, $k=\dfrac{9}{4}$.

The tangents at the extremities of a focal chord of $y^2=12x$ meet on the directrix $x+3=0$.

1. Write the parabola in standard form $y^2=4ax$; comparing with $y^2=12x$ gives $4a=12$, so $a=3$.
2. A well-known property of the parabola $y^2=4ax$ is that the tangents drawn at the two endpoints of any focal chord always intersect on the directrix.
3. The directrix of $y^2=4ax$ is the line $x=-a$.
4. Substituting $a=3$: the directrix is $x=-3$, i.e. $x+3=0$.
5. Hence the point of intersection of these tangents always lies on the line $x+3=0$.

Using the standard result that tangents to $y^2=4ax$ at parameters $t_1,t_2$ meet at $(at_1t_2,\,a(t_1+t_2))$, with $a=2$, $t_1=t$, $t_2=3t$ gives the intersection point $(6t^2,8t)$.

1. Compare $y^2=8x$ with the standard form $y^2=4ax$: $4a=8\Rightarrow a=2$.
2. A point on the parabola with parameter $t_i$ is $(at_i^2,2at_i) = (2t_i^2,4t_i)$, and the tangent there is $t_iy = x + at_i^2 = x + 2t_i^2$.
3. Tangent at $t_1=t$: $ty = x + 2t^2$ ... (I)
4. Tangent at $t_2=3t$: $3ty = x + 2(3t)^2 = x+18t^2$ ... (II)
5. Subtract (I) from (II): $2ty = 16t^2 \Rightarrow y = 8t$ (for $t\ne0$).
6. Substitute back into (I): $x = ty - 2t^2 = t(8t) - 2t^2 = 8t^2-2t^2=6t^2$.
7. So the intersection point is $(6t^2, 8t)$, matching the general formula $(at_1t_2, a(t_1+t_2)) = (2\cdot t\cdot3t,\ 2(4t)) = (6t^2,8t)$.

Computing the foot of the perpendicular from a focus of the hyperbola onto a general tangent line shows the locus is the auxiliary circle $x^2+y^2=a^2$.

1. A tangent to $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ in slope form is $y=mx+c$ with the tangency condition $c^2=a^2m^2-b^2$, i.e. $mx-y+c=0$.
2. Take the focus $S=(ae,0)$, where $b^2=a^2(e^2-1)$.
3. Using the perpendicular-foot formula for a point $(x_0,y_0)$ onto $ux+vy+w=0$ (here $u=m,\,v=-1,\,w=c$): with $k=\dfrac{ame+c}{m^2+1}$, the foot is $x=ae-mk,\ y=k$.
4. Compute $x^2+y^2=(ae-mk)^2+k^2 = a^2e^2-2aemk+k^2(m^2+1)$.
5. Since $k(m^2+1)=ame+c$, this simplifies to $a^2e^2 + k(c-ame)$, and substituting $k=\dfrac{ame+c}{m^2+1}$ gives $x^2+y^2=a^2e^2+\dfrac{c^2-a^2m^2e^2}{m^2+1}$.
6. Using $c^2=a^2m^2-b^2=a^2m^2-a^2(e^2-1)$, the numerator becomes $a^2(1-e^2)(m^2+1)$, so the fraction reduces to $a^2(1-e^2)$.
7. Hence $x^2+y^2=a^2e^2+a^2(1-e^2)=a^2$ — independent of $m$, so the locus is the full circle $x^2+y^2=a^2$, the auxiliary circle of the hyperbola.

Use the standard normal-chord relation for the parabola $y^2=4ax$: if the normal at parameter $t_1$ re-meets the curve at $t_2$, then $t_2=-t_1-\frac{2}{t_1}$; rearranging gives the required expression.

1. Points on $y^2=4ax$ are $P(t)=(at^2,2at)$.
2. Slope of tangent at $t_1$ is $\dfrac{dy}{dx}=\dfrac{1}{t_1}$, so slope of the normal at $t_1$ is $-t_1$.
3. Equation of the normal at $t_1$: $y - 2at_1 = -t_1(x-at_1^2)$, i.e. $y = -t_1x + 2at_1 + at_1^3$.
4. This normal meets the parabola again at parameter $t_2$, i.e. at $(at_2^2, 2at_2)$. Substituting: $2at_2 = -t_1(at_2^2) + 2at_1+at_1^3$.
5. The well-known result obtained by solving this (equivalent to using that a normal chord's parameters satisfy $t_2 = -t_1 - \dfrac{2}{t_1}$) is: $t_2 = -t_1 - \frac{2}{t_1}.$
6. Rearranging: $t_1 + \dfrac{2}{t_1} = -t_2$.
7. This matches option (a).

The director circle of the given ellipse has radius $5$.

1. Divide by $144$: $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$, so $a^2=16,\ b^2=9$.
2. The director circle (locus of the point of intersection of perpendicular tangents) of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ is $x^2+y^2=a^2+b^2$.
3. Radius $=\sqrt{a^2+b^2}=\sqrt{16+9}=\sqrt{25}=5$.
4. Options (a) $\sqrt7$, (b) $4$, (c) $3$ correspond to $\sqrt{a^2-b^2}$, $a$, and $b$ respectively — not the director-circle radius.

Perpendicular tangents to a parabola always meet on its directrix.

1. For the parabola $y^2=4ax$, a tangent in slope form is $y=mx+\dfrac{a}{m}$ (for slope $m\ne0$).
2. Two tangents with slopes $m_1$ and $m_2$ satisfying the perpendicularity condition $m_1m_2=-1$ intersect where $m_1x+\dfrac{a}{m_1}=m_2x+\dfrac{a}{m_2}$.
3. Solving this pair of tangent equations for the intersection point and eliminating $m_1,m_2$ using $m_1m_2=-1$ shows the $x$-coordinate of every such intersection point is the constant $x=-a$, regardless of $y$.
4. $x=-a$ is precisely the equation of the directrix of $y^2=4ax$.
5. Options (a), (c), (d) name other special lines of the parabola (latus rectum $x=a$, tangent at the vertex $x=0$, axis $y=0$), none of which is the correct locus.