Question

Let image of a variable point $(–4\cos \theta ,–3\sin \theta )$ about the line $x+y=0$ lie on curve $C$ and pair of straight lines represented by the equation $m(x^2–y^2)+(m^2–1)xy=0$ denotes two chords of the curve $C$ for $m\in R-\left\{0\right\}.$ Then choose the correct option(s).

• A. If perpendiculars are drawn on the lines joining end points of the chords from the origin, then locus of feet of these perpendiculars is $x^2+y^2=\frac{144}{25}.$
• B. Minimum length of the segment joining end points of the given pair of chords is $\frac{24}{5}.$
• C. Product of the length of perpendiculars drawn from the point $(0,\sqrt{7})$ and $(0,-\sqrt{7})$ upon a variable tangent to curve $C$ is $9.$
• D. Product of the length of perpendiculars drawn from the point $(0,\sqrt{7})$ and $(0,-\sqrt{7})$ upon a variable tangent to curve $C$ is $25.$

Answer 1

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

The correct options are
A If perpendiculars are drawn on the lines joining end points of the chords from the origin, then locus of feet of these perpendiculars is $x^2+y^2=\frac{144}{25}.$
B Minimum length of the segment joining end points of the given pair of chords is $\frac{24}{5}.$
C Product of the length of perpendiculars drawn from the point $(0,\sqrt{7})$ and $(0,-\sqrt{7})$ upon a variable tangent to curve $C$ is $9.$

As we know that image of any point $(a,b)$ about the line $y=-x$ is $(-b,-a),$
$\therefore$ Image of $(-4\cos \theta ,-3\sin \theta )$ in $y=-x$ is $(3\sin \theta ,4\cos \theta )$
$\therefore$ Locus of this point is
$h=3\sin \theta ,k=4\cos \theta$
$\Rightarrow \frac{x^2}{9}+\frac{x^2}{16}=1$

Given pair of chords
$m(x^2-y^2)+(m^2-1)xy=0$
$\Rightarrow m{x}^2+m^{2}xy-xy-m{y}^2=0$
$\Rightarrow mx(x+my)-y(x+my)=0)$
$\Rightarrow (x+my)(mx-y)=0$
$\Rightarrow y=mx$ and $y=\frac{-1}{m}x$
$\therefore$ Given chords are perpendicular and passing through the origin.

Length of the perpendicular from the centre on the line joining end points of perpendiculars is always constant and which is given by,
$\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$


$\Rightarrow p=\frac{ab}{\sqrt{a^2+b^2}}=\frac{3\cdot 4}{5}=\frac{12}{5}$
$\therefore$ Locus of foot of the perpendicular is
$x^2+y^2=\frac{144}{25}$

$PQ=p\cot \varphi +p\tan \varphi =p(\tan \varphi +\cot \varphi )$
$PQ$ will be minimum if $\varphi ={45}^{\circ }$
$\therefore PQ{|}_{min}=2p=\frac{24}{5}$


$(0,\sqrt{7})$ and $(0,-\sqrt{7})$ are the foci of ellipse $\frac{x^2}{9}+\frac{y^2}{16}=1$
Hence, $p_1{p}_2=b^2=9$