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 $