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Hyperbola

Section: Mathematics

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Question : 72 of 98

Marks: +1, -0

A line parallel to the straight line

2x - y = 0

istangent to the hyperbola

\frac{x^2}{4} - \frac{y^2}{2} = 1

(x_1, y_1).

x_1^2 + 5 y_1^2

[2 Sep 2020 Shift 1]

5
6
8
10

Solution:

Slope of tangent is

2,

Tangent of hyperbola

\frac{x^2}{4} - \frac{y^2}{2} = 1

at the point

(x_1, y_1)

is

\frac{x x_1}{4} - \frac{y y_1}{2} = 1 \;\; (T=0)

Slope

: \frac{1}{2} \frac{x_1}{y_1} = 2 \Rightarrow x_1 = 4 y_1

....(1)

(x_1, y_1)

lies on hyperbola

\Rightarrow \sqrt{ \frac{x_1^2}{4} - \frac{y_1^2}{2} } = 1

...(2)

From (1) & (2)

\frac{(4 y_1)^2}{4} - \frac{y_1^2}{2} = 1 \Rightarrow 4 y_1^2 - \frac{y_1^2}{2} = 1

\Rightarrow 7 y_1^2 = 2

\Rightarrow y_1^2 = \frac{2}{7}

Now

x_1^2 + 5 y_1^2 = (4 y_1)^2 + 5 y_1^2

= (21) y_1^2 = 21 \times \frac{2}{7} = 6

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