"Let the normal at a point P on the curve y2 – 3x2 + y + 10 = 0 intersect the y-axis at $\left( {0,{3 \over 2}} \right)$ . If m is the slope of the tangent at P to the curve, then |m| is equal to"

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Accepted Answer

"Given curve : y² – 3x² + y + 10 = 0

$\Rightarrow$ 2y${{dy} \over {dx}}$ - 6x + ${{dy} \over {dx}}$ = 0

$\Rightarrow$ ${{dy} \over {dx}}$ = ${{6x} \over {2y + 1}}$

Let P be (x₁, y₁)

Slope of tangent at P = ${{6{x_1}} \over {2{y_1} + 1}}$

$\therefore$ Slope of normal at P = $- {{2{y_1} + 1} \over {6{x_1}}}$

$\Rightarrow$ Equation of normal (y – y₁) = $- \left( {{{2{y_1} + 1} \over {6{x_1}}}} \right)$(x – x₁)

This normal passes through point $\left( {0,{3 \over 2}} \right)$.

$\therefore$ (${{3 \over 2}}$ – y₁) = $- \left( {{{2{y_1} + 1} \over {6{x_1}}}} \right)$(0 – x₁)

$\Rightarrow$ y₁ = 1

Put y₁ = 1 in equation of curve , then we get x₁ = $\pm$2

$\Rightarrow$ |m| = slope of tangent = $\left| {{{6{x_1}} \over {2{y_1} + 1}}} \right|$ = ${{12} \over 3}$ = 4"

Let the normal at a point P on the curve
y² – 3x² + y + 10 = 0 intersect the y-axis at $\left( {0,{3 \over 2}} \right)$ .
If m is the slope of the tangent at P to the curve, then |m| is equal to

Solution

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Let the normal at a point P on the curve y2 – 3x2 + y + 10 = 0 intersect the y-axis at $\left( {0,{3 \over 2}} \right)$ . If m is the slope of the tangent at P to the curve, then |m| is equal to

Solution

About this question

More Application of Derivatives questions

Drill 25 more like these. Every day. Free.

Let the normal at a point P on the curve
y² – 3x² + y + 10 = 0 intersect the y-axis at $\left( {0,{3 \over 2}} \right)$ .
If m is the slope of the tangent at P to the curve, then |m| is equal to