## 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

Ans: (b)
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.

## A tangent of the ellipse x^2/a^2 + y^2/b^2 = 1 is normal to the hyperbola x^2/4 – y^2/1 = 1 and

Q: A tangent of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is normal to the hyperbola $\frac{x^2}{4} – \frac{y^2}{1} = 1$ and it has equal intercepts with positive x and y axes, then the value of a2 + b2 is

(A) 5

(B) 25

(C) 16

(D) 25/9

Sol. The equation of normal to the hyperbola $\frac{x^2}{4} – \frac{y^2}{1} = 1$ at (2 sec θ , tan θ) is 2x cos θ + y cot θ = 5

Slope of normal = – 2 sin θ = – 1

⇒ θ = π/6

y-intercept of normal = 5/cotθ = 5/√3

Since it touches the ellipse $\frac{x^2}{a^2} + \frac{y^2}{a^2} = 1$

a2 + b2 = 25/9

## The equation of common tangents to the ellipse x^2 + 2y^2 = 1 and the circle x^2 + y^2 = 2/3 is

Q: The equation of common tangents to the ellipse x2 + 2y2 = 1 and the circle x2 + y2 = 2/3 is

(A) $\large y = \frac{\sqrt{7}}{\sqrt{2}} x + \sqrt{3}$

(B) y = 7x + √3

(C) y = √7x + √2

(D) none of these

Sol. The equation of any tangent to the circle x2 + y2 = 2/3 is $\large y = \pm \frac{\sqrt{2}}{\sqrt{3}} \sqrt{1 + m^2}$ since it touches the given ellipse then

$\large \frac{\sqrt{2}}{\sqrt{3}} \sqrt{1 + m^2} = \sqrt{m^2 – \frac{1}{2}}$

Squaring ,

$\large \frac{2}{3} + \frac{2}{3}m^2 = m^2 -\frac{1}{2}$

$\large m = \pm \frac{\sqrt{7}}{\sqrt{2}}$

$\large y = \frac{\sqrt{7}}{\sqrt{2}} x + \sqrt{3}$ is the common tangent.

## If the line y = x + √3 touches the ellipse  $\frac{x^2}{4} + \frac{y^2}{1} = 1$ , then the point of contact is

Q: If the line y = x + √3 touches the ellipse  $\large \frac{x^2}{4} + \frac{y^2}{1} = 1$ , then the point of contact is

(A) $(\frac{2}{\sqrt{3}} , \frac{1}{\sqrt{3}})$

(B) $(- \frac{4}{\sqrt{3}} , \frac{1}{\sqrt{3}})$

(C) $(-\frac{2}{\sqrt{3}} , -\frac{1}{\sqrt{3}})$

(D) none of these

Sol. Let the point of contact be (x’, y’) then equation of tangent is $\large \frac{x x’}{4} + \frac{y y’}{1} = 1$ comparing it with y – x = √3 , we get

$\large \frac{x’}{-4} = \frac{y’}{1} = \frac{1}{\sqrt{3}}$

$\large x’ = \frac{-4}{\sqrt{3}}$ and $\large y’ = \frac{1}{\sqrt{3}}$

## The minimum area of triangle formed by the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and coordinate axes is

Q: The minimum area of triangle formed by the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and coordinate axes is

(A) ab sq. units

(B) $\frac{a^2 + b^2}{2}$ sq. units

(C) $\frac{(a + b)^2}{2}$ sq. units

(D) $\frac{a^2 + ab + b^2}{3}$ sq. units

Sol. A tangent of the given ellipse is $\large y = m x + \sqrt{a^2 m^2 + b^2}$

It meets the axes at $\large (\frac{-\sqrt{a^2 m^2 + b^2}}{m} , 0 )$ and $(0 , \sqrt{a^2 m^2 + b^2})$

Hence the area of the triangle is $\large \frac{1}{2}|\frac{a^2 m^2 + b^2}{m}|$

$\large = \frac{1}{2}|a^2 m + \frac{b^2}{m}| \ge a b$

## If PQR be an equilateral triangle in the auxiliary circle of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (a > b) and

Q:If PQR be an equilateral triangle in the auxiliary circle of the ellipse $\large \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (a > b) and P’Q’R’ be corresponding triangle inscribed within the ellipse then centroid of the triangle P’Q’R’ lies at

(A) centre of the ellipse

(B) focus of the ellipse

(C) vertex of the ellipse

(D) none of these

Sol. Let P(θ), Q(θ + 2π/3) and R(θ + 4π/3)  , then centroid of the triangle P’Q’R’ ∈ (x, y) is

$\large x = \frac{a(cos\theta + cos(\theta + \frac{2\pi}{3})) + cos(\theta + \frac{4\pi}{3}))}{3} = 0$

$\large y = \frac{b(sin\theta + sin(\theta + \frac{2\pi}{3})) + sin(\theta + \frac{4\pi}{3}))}{3} = 0$