SATHEE: Hyperbola 2 Question 1

Hyperbola 2 Question 1

1. The equation of a common tangent to the curves, $y^{2} = 16 x$ and $x y = - 4$ , is

(2019 Main, 12 April II)

(a) $x - y + 4 = 0$

(b) $x + y + 4 = 0$

(d) $2 x - y + 2 = 0$

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Answer:

Correct Answer: 1. (a)

Solution:

Key Idea An equation of tangent having slope

’ $m$ ’ to parabola $y^{2} = 4 a x$ is $y = m x + \frac{a}{m}$ .

Given equation of curves are

$y^{2} = 16 x (parabola)$

and $x y = - 4$ (rectangular hyperbola)

Clearly, equation of tangent having slope ’ $m$ ’ to parabola

(i) is $y = m x + \frac{4}{m}$

Now, eliminating $y$ from Eqs. (ii) and (iii), we get

$x m x + \frac{4}{m} = - 4 \Rightarrow m x^{2} + \frac{4}{m} x + 4 = 0$

which will give the points of intersection of tangent and rectangular hyperbola.

Since, line $y = m x + \frac{4}{m}$ is also a tangent to the rectangular hyperbola.

$∴$ Discriminant of quadratic equation $m x^{2} + \frac{4}{m} x + 4 = 0$ , should be zero.

[ $∵$ there will be only one point of intersection]

$\begin{array}{ll} \Rightarrow & D = {\frac{4}{m}}^{2} - 4 (m) (4) = 0 \\ \Rightarrow & m^{3} = 1 \Rightarrow m = 1 \end{array}$

So, equation of required tangent is $y = x + 4$ .

Hyperbola 2 Question 1

1. The equation of a common tangent to the curves, $y^{2} = 16 x$ and $x y = - 4$ , is

Answer:

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Hyperbola 2 Question 1

1. The equation of a common tangent to the curves, y2=16x and xy=−4, is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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1. The equation of a common tangent to the curves, $y^{2} = 16 x$ and $x y = - 4$ , is