If y=4x−5 is a tangent to the curve y2=px3+q at (2,3) then pq=

Solution:

Point (2, 3) lies on $y^{2} = p x^{3} + q .$ . Therefore,

$9 = 8 p + q$

Now,

$\begin{array}{l} y^{2} = p x^{3} + q \\ \Rightarrow 2 y \frac{d y}{d x} = 3 p x^{2} \\ \Rightarrow \frac{d y}{d x} = \frac{3 p x^{2}}{2 y} \Rightarrow {(\frac{d y}{d x})}_{(2, 3)} = \frac{12 p}{6} = 2 p . \end{array}$

Since , $y = 4 x - 5$ is $y^{2} = p x^{3} + q at (2, 3)$

$∴ {(\frac{d y}{d x})}_{(2, 3)} =$ (Slope of the line $y = 4 x - 5)$

$\Rightarrow 2 p = 4 \Rightarrow p = 2$

Putting $p = 2 i n (i)$ , we get $q = - 7$

Hence, $p q = - 14$

If $y = 4 x - 5$ is a tangent to the curve $y^{2} = p x^{3} + q at (2, 3)$ then $p q =$

Solution:

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