Question

Find the equation of the ellipse satisfying the given condition $e=\frac{3}{4}$, foci on Y-axis, centre at origin and passes through (6,4).

Or
Find the equation of the hyperbola with vertices at $(\pm 5,0)$ and foci $(\pm 7,0)$

Answer 1

Let the equation of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\dots \left(i\right)$

If the ellipse passes through (6,4) then
$\frac{(6{)}^2}{a^2}+\frac{(4{)}^2}{b^2}=1\Rightarrow \frac{36}{a^2}+\frac{16}{b^2}=11\dots \left(ii\right)$

We know that, $e=\sqrt{1-\frac{a^2}{b^2}}[\because b>a]$

$\Rightarrow \frac{3}{4}=\sqrt{1-\frac{a^2}{b^2}}\Rightarrow \frac{9}{16}=1-\frac{a^2}{b^2}$

$\Rightarrow \frac{a^2}{b^2}=1-\frac{9}{16}=\frac{7}{16}\Rightarrow a^2=\frac{7}{16}{b}^{2}1\dots \left(iii\right)$

From Eqs. (i) and (iii), we get

$\frac{36\times 16}{7b^2}+\frac{16}{b^2}=1\Rightarrow \frac{16}{7b^2}(36+7)=1$

$\Rightarrow \frac{1}{b^2}=\frac{7}{16\times 43}\Rightarrow b^2=\frac{16\times 43}{7}$

Put the value of b^2 in Eq. (iii) we get

$a^2=\frac{7}{16}\times \frac{16\times 43}{7}=43$

On substituting the values of $a^2$ and $b^2$ in Eq. (i), we get

$\frac{x^2}{43}+\frac{7y^2}{16\times 43}=1$

$\therefore 16x^2+7y^2=688$, which is the required equation of ellipse.

Or

Since, the vertices are on the X-axis with origin at the mid-point.
$\therefore (\pm a,0)=(\pm 5,0)and(\pm c,0)=(\pm 7,0)$

Then, equation of the hyperbola is of the form
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Since, the vertices are $(\pm 5,0)$ and foci are $(\pm 7,0])$
$\therefore a=5andc=7$

We know that, in hyperbola
$c^2=a^2+b^2$

$\Rightarrow 7^2=5^2+b^2\Rightarrow b^2=49-25=24$

On putting the value of $b^2$ in Eq. (i) we get
$\frac{x^2}{25}-\frac{y^2}{24}=1\Rightarrow 24x^2-25y^2=600$