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MHTCET 14 May 2023 Shift 2 Solved Paper

Section: Mathematics

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Question : 66 of 150

Marks: +1, -0

If

y=4x-5

is a tangent to the curve

y^2=p x^3+q

(2,3)

p-q

is

$-5$
$5$
$9$
$-9$

Solution:

y^2=p x^3+q

...(i)

Differentiating both sides w.r.t.

x

, we get

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

\Rightarrow \frac{dy}{dx}=\frac{3p}{2} \left(\frac{x^2}{y}\right)

\therefore \left(\frac{dy}{dx}\right)_{(2,3)}=\frac{3p}{2} \times \frac{4}{3}=2p

Slope of the line

y=4x-5

is 4 .

Since the line touches the curve, their slopes are equal.

\therefore 2p=4 \Rightarrow p=2

Since

(2,3)

lies on

y^2=p x^3+q

.

\therefore 2p=4 \Rightarrow p=2

Since

(2,3)

lies on

y^2=p x^3+q

.

\therefore 9=2 \times 8+q \Rightarrow q=-7

\therefore p-q=2+7=9

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