SATHEE: Parabola 2 Question 15

Parabola 2 Question 15

15. Equation of common tangent of $y = x^{2}, y = - x^{2} + 4 x - 4$ is

$(2006, 5 M)$

(a) $y = 4 (x - 1)$

(b) $y = 0$

(d) $y = - 30 x - 50$

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Let $a, r, s, t$ be non-zero real numbers. Let $P (a t^{2}, 2 a t), Q, R (a r^{2}, 2 a r)$ and $S (a s^{2}, 2 a s)$ be distinct point on the parabola $y^{2} = 4 a x$ . Suppose that $P Q$ is the focal chord and lines $Q R$ and $P K$ are parallel, where $K$ is point $(2 a, 0)$ .

(2014, Adv.)

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Solution:

The equation of tangent to $y = x^{2}$ , be $y = m x - \frac{m^{2}}{4}$ . Putting in $y = - x^{2} + 4 x - 4$ , we should only get one value of $x$ i.e. Discriminant must be zero.

$\begin{aligned} ∴ m x - \frac{m^{2}}{4} & = - x^{2} + 4 x - 4 \\ \Rightarrow x^{2} + x (m - 4) + 4 - \frac{m^{2}}{4} & = 0 \\ D & = 0 \end{aligned}$

Now, $(m - 4)^{2} - (16 - m^{2}) = 0$

$\Rightarrow 2 m (m - 4) = 0 \Rightarrow m = 0, 4$

$∴ y = 0$ and $y = 4 (x - 1)$ are the required tangents.

Hence, (a) and (b) are correct answers.

Parabola 2 Question 15

15. Equation of common tangent of $y = x^{2}, y = - x^{2} + 4 x - 4$ is

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Solution:

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Parabola 2 Question 15

15. Equation of common tangent of y=x2,y=−x2+4x−4 is

Passage Based Problems

Passage

Solution:

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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15. Equation of common tangent of $y = x^{2}, y = - x^{2} + 4 x - 4$ is