Question

Given ellipse $\frac{x^2}{16}+\frac{y^2}{7}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$, if the ordinate of one of the points of intersection is produced to cut an asymptote at $P$, then which of the following is true?

• A. they have the same foci
• B. square of the ordinate of point of intersection is $\frac{63}{25}$.
• C. sum of the square of coordinates of $P$ is 16
• D. P lies on the auxiliary circle formed by ellipse.

Answer 1

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

$\frac{x^2}{16}+\frac{y^2}{7}=1\dots (1)$
$\Rightarrow a^2=16,b^2=7$
i.e. $a=4,b=\sqrt{7}$
$\therefore e^2=\frac{a^2-b^2}{a^2}\Rightarrow e=\frac{3}{4}$
$\therefore \text{ foci }\equiv (\pm ae,0)=(\pm 3,0)$
$\frac{x^2}{(144/25)}-\frac{y^2}{(81/25)}=1\dots ..(2)$
$\Rightarrow a^2=\frac{144}{25},b^2=\frac{81}{25}$
i.e. $a=\frac{12}{5},b=\frac{9}{5}$
$\therefore e^2=\frac{a^2+b^2}{a^2}\Rightarrow e=\frac{5}{4}$
$\text{ foci }\equiv (\pm ae,0)=(\pm 3,0)$
solving (1) and (2) we get
$y^2=\frac{63}{25}\Rightarrow y=\pm \frac{3\sqrt{7}}{5}\Rightarrow x=\pm \frac{16}{5}$
one of the point of intersection is $\left(\frac{16}{5},\frac{3\sqrt{7}}{5}\right)$
The equation of the asymptote is $\frac{x^2}{144}-\frac{y^2}{81}=0$
The abscissa of $P$ is $\frac{16}{5}$
Its ordinate is given by $\frac{y^2}{81}=\frac{16\times 16}{25\times 144}$
$\therefore y=\pm \frac{12}{5}$
$\therefore P\equiv \left(\frac{16}{5},\frac{12}{5}\right)\Rightarrow {\left(\frac{16}{5}\right)}^2+{\left(\frac{12}{5}\right)}^2=16$
Equation of the auxiliary circle formed on major axis of ellipse $x^2+y^2=16$ P lies on it.