Question

Consider the curves $C_1:\frac{x^2}{9}+\frac{y^2}{4}=1,$ $C_2:(x+4{)}^2+y^2=1$ and $C_3:y^2=4a(x-3)$. If all three curves have two common tangents, then

• A. Length of latus rectum of curve $C_3$ is $\frac{24}{7}$ units.
• B. $x-$intercept of common tangents is $-\frac{23}{3}$.
• C. Area of the triangle formed by the common tangents and $y-$axis is $\frac{529}{12\sqrt{7}}$ sq. units.
• D. Product of the length of the perpendiculars drawn from $(-\sqrt{5},0)$ and $(\sqrt{5},0)$ to any of the common tangents is equal to $4$.

Answer 1

Answer: (A), (B), (C), (D)

Product of the length of the perpendiculars drawn from $(-\sqrt{5},0)$ and $(\sqrt{5},0)$ to any of the common tangents is equal to $4$.

Given curves are $C_1:\frac{x^2}{9}+\frac{y^2}{4}=1,C_2:(x+4{)}^2+y^2=1$ and $C_3:y^2=4a(x-3)$


Tangent to ellipse $C_1$ in slope form is
$y=mx\pm \sqrt{9m^2+4}$
This is a tangent to the circle $C_2$, so distance from the centre is equal to the radius.
$\frac{|-4m\pm \sqrt{9m^2+4}|}{\sqrt{1+m^2}}=1\Rightarrow m=\pm \frac{3}{4\sqrt{7}}$
Equation of tangents will be
$y=\frac{3}{4\sqrt{7}}x+\frac{23}{4\sqrt{7}}$ or $y=-\frac{3}{4\sqrt{7}}x-\frac{23}{4\sqrt{7}}$

Tangent to parabola will be of the form $y=m(x-3)+\frac{a}{m}$
On comparing, $a=\frac{6}{7}$
Length of latus rectum $=4a=\frac{24}{7}$
$x-$ intercept $=-\frac{23}{3}$

Area of triangle $=2\times \frac{1}{2}\times \frac{23}{3}\times \frac{23}{4\sqrt{7}}=\frac{529}{12\sqrt{7}}$

Here, $(\sqrt{5},0)$ and $(-\sqrt{5},0)$ are foci of the given ellipse and the product of lengths of perpendicular from foci upon any tangent is the square of semi-minor axis $=b^2=4$