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)

$\frac{x^2}{16}+\frac{y^2}{7}=1\ldots(1)$
$ \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} $
$\therefore \text { foci } \equiv( \pm a e, 0)=( \pm 3,0)$
$\frac{x^2}{(144 / 25)}-\frac{y^2}{(81 / 25)}=1\ldots . .(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.