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KEAM 2013 Math Paper

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Question : 101 of 120

Marks: +1, -0

If

y = 4x-5

is a tangent to the curve

y^2 = p x^3 + q

(2,3),

p+q=

$-5$
5
$-9$
9
0

Solution:

Given curve is

y^{2} = p x^{3} + q

....(i)

On differentiating w.r.t.

x

, we get

2 y \frac{dy}{dx} = 3 p x^{2} \Rightarrow \frac{dy}{dx} = \frac{3 p x^{2}}{2 y}

At

(2,3), \left(\frac{dy}{dx}\right)_{(2,3)} = \frac{12 p}{6} = 2 p

since, line

y = 4x - 5

is a tangent to the given curve.

\therefore

Slope

=4=2 p

\Rightarrow p=2

Now, put the value of

p, x

and

y

in Eq. (i), we get

(3)^{2} = (2)(2)^{3} + q

\Rightarrow 9 = 16 + q

\Rightarrow q = -7

Hence,

p+q = 2-7 = -5

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