# Ellipse

Co-ordinate Geometry

## Equation of tangent to two ellipse x^2/9 + y^2/4 = 1 , which cut off equal intercepts on the axes is

Q: Equation of tangent to two ellipse $\large \frac{x^2}{9} + \frac{y^2}{4} = 1$ which cut off equal intercepts on the axes is

(A) $y = x + \sqrt{13}$

(B) $y = -x + \sqrt{13}$

(C) $y = x – \sqrt{13}$

(D) $y = -x – \sqrt{13}$

Sol. Equation of tangent to ellipse $\large \frac{x^2}{9} + \frac{y^2}{4} = 1$ is

$\large y = mx \pm \sqrt{9m^2 + 4}$

Here m = ± 1

⇒ tangents are $y = \pm x \pm \sqrt{13}$

## A man running a race course notes that the sum of his distances from the two flag posts is always 10m and the distance between the flag posts is 8m.

Q: A man running a race course notes that the sum of his distances from the two flag posts is always 10m and the distance between the flag posts is 8m. The equation of the path traced by him is

(A) $\large \frac{x^2}{25} + \frac{y^2}{9} = 1$

(B) $\large \frac{x^2}{4} + \frac{y^2}{1} = 1$

(C) $\large \frac{x^2}{2} + \frac{y^2}{1} = 1$

(D) none of these

Sol. Here 2a = 10 ⇒  a = 5 and 2ae = 8

⇒ e = 8/10 = 4/5

$\large b^2 = 25(1-\frac{16}{25}) = 9$

⇒  $\large \frac{x^2}{25} + \frac{y^2}{9} = 1$ is equation of the path.

## A square is inscribed inside the ellipse x^2/a^2 + y^2/b^2 = 1 , the length of the side of the square is

Q: A square is inscribed inside the ellipse $\large \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ , the length of the side of the square is

(A) $\frac{ab}{\sqrt{a^2 + b^2}}$

(B) $\frac{2ab}{\sqrt{a^2 + b^2}}$

(C) $\sqrt{a^2 + b^2}$

(D) None of these

Sol.

Let one vertex of the square be (p, p) then

$\large \frac{p^2}{a^2} + \frac{p^2}{b^2} = 1$

$\large p^2 = \frac{a^2 b^2}{a^2 + b^2}$

⇒ $\large p = \frac{ab}{\sqrt{a^2 + b^2}}$

⇒  Length of side $\large = \frac{2ab}{\sqrt{a^2 + b^2}}$

## The centre of the ellipse 3x^2 + 4y^2 – 12x – 8y + 4 = 0 is

Q: The centre of the ellipse 3x2 + 4y2 – 12x – 8y + 4 = 0 is

(A) (1, 1)

(B) (2, 1)

(C) (3, 1)

(D) none of these

Sol. The equation of the ellipse can be written as

3 (x2 – 4x + 4) + 4 (y2 – 2y + 1) = 12

⇒ $\large \frac{(x-2)^2}{4} + \frac{(y-1)^2}{3} = 1$

⇒ (2, 1) is the centre.

## The equation of the ellipse with e =3/4  , foci on y-axis, centre of the origin and passing through the point (6, 4) is

Q: The equation of the ellipse with e =3/4  , foci on y-axis, centre of the origin and passing through the point (6, 4) is

(A) x2 + 2y2 = 16

(B) 16x2 + 7y2 = 688

(C) 16x2 + 7y2 = 344

(D) none of these

Sol: $\large b^2 = a^2 (1-\frac{9}{16}) = \frac{7}{16}a^2$

Then equation is $\large \frac{16 x^2}{7 a^2} + \frac{y^2}{a^2} = 1$ , it passes through (6, 4)

$\large \frac{16 \times 6^2}{7 a^2} + \frac{16}{b^2} = 1$

$\large a^2 = \frac{688}{7}$

16x2 + 7y2 = 688 is ellipse.