JAMIA 2021 — Mathematics PYQ

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Question

JAMIA 2021 — Mathematics PYQ

JAMIA | Mathematics | 2021

If slope of common tangent to curves $4 x^{2} + 9 y^{2} = 36$ and 4 $x^{2}$ + 4 $y^{2}$ = 31 is m, then $m^{2}$ is equal to:

Choose the correct answer:

Explanation

Curve 1: Ellipse

\frac{4x^2}{36} + \frac{9y^2}{36} = 1 \implies \frac{x^2}{9} + \frac{y^2}{4} = 1

Here, $a^2 = 9$ and $b^2 = 4$ .

The equation of a tangent with slope $m$ is:

y = mx \pm \sqrt{a^2m^2 + b^2}

y = mx \pm \sqrt{9m^2 + 4} \quad \dots (1)

Curve 2: Circle

4x^2 + 4y^2 = 31 \implies x^2 + y^2 = \frac{31}{4}

The radius squared is $r^2 = \frac{31}{4}$ .

The equation of a tangent with slope $m$ is:

y = mx \pm r\sqrt{1 + m^2}

y = mx \pm \sqrt{\frac{31}{4}(1 + m^2)} \quad \dots (2)

Condition for Common Tangent

Equating the constant terms from (1) and (2):

9m^2 + 4 = \frac{31}{4}(1 + m^2)

4(9m^2 + 4) = 31(1 + m^2)

36m^2 + 16 = 31 + 31m^2

36m^2 - 31m^2 = 31 - 16

5m^2 = 15

m^2 = 3

Final Answer:

m^2 = 3