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Solved Examples on Ellipse

Illustration 1: Tangent is drawn to ellipse x2/27 + y2 = 1 at (3√3cos θ, sin θ) (where θ ∈ (0, π/2)). Then what is the value of θ for which the sum of intercepts on axis made by this tangent is minimum?

Solution: The given equation of parabola is x2/27 + y2 = 1

A tangent is drawn at the point (3√3cos θ, sin θ).

Hence, the equation of tangent at this point is given by

(x cos θ)/ 3√3 + (y sin θ)/ 1 = 1

Thus, the sum of intercepts = [3√3/ (cos θ) + 1/ (sin θ)] = f (θ) (say)

Hence, f’ (θ) = (3√3 sin3θ – cos3θ)/ (sin2θ – cos2θ)

Now, equating f’ (θ) = 0, we get

sin3θ = cos3θ/33/2

Hence, tan θ = 1/√3, i.e. θ = π/6

And so, at θ = π/6, f” (0) > 0

Hence, this implies that the tangent is minimum at π/6.

Illustration 2: Prove that, in an ellipse, the perpendicular from a focus upon any tangent and the line joining the center of the ellipse of the point of contact meet on the corresponding directrix. (2002)

Solution: Let the equation of the ellipse be x2/a2 + y2/b2 = 1

Then any point on the ellipse is of the form P (a cos θ, b sin θ).

The equation of tangent at point P is given by

(x cos θ) /a + (y sin θ) /b = 1

The equation of line perpendicular to tangent is

(x sin θ) /b - (y cos θ) /a = μ

Now, since this passes through the focus (ae, 0), so

(ae sin θ)/b – 0 = μ

This gives μ = (ae sin θ)/b

Therefore the equation is (x sin θ) /b - (y cos θ) /a = (ae sin θ)/b    …. (1)

Equation of the line joining center and point of contact P (a cos θ, b sin θ) is

y = b/a (tan θ)x                                                                                       … (2)

Point of intersection Q of equations (1) and (2) has x coordinate a/e.

Hence, Q lies on the corresponding directrix x = a/e.

Illustration 3: Find the equation of the common tangent in 1st quadrant to the circle x2 + y2 = 16 and the ellipse x2/25 + y2/ 4 = 1. Also find the length of the intercept of the tangent between the coordinate axis. (2005)

Solution: Let the common tangent to x2 + y2 = 16 and x2/25 + y2/ 4 = 1 be

y = mx + 4 √(1 + m2)

And y = mx + √(25m2 + 4)

Since both these represent the same tangent, so

4 √(1+m2) = √(25m2 + 4)

So, 16 (1 + m2) = 25m2 + 4

Hence, 9m2 = 12

Which gives m = ± 2/√3

Since, tangent is in first quadrant, so m < 0

Hence, m = -2/√3

So, the equation of the common tangent is y...

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