1e:I[80107,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1f:I[76763,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1d:T42c,{"@context":"https://schema.org","@graph":[{"@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https://www.divinejee.com"},{"@type":"ListItem","position":4,"name":"Let the normal at a point P on the curve y2 3x2 y 10 0 inter","item":"https://www.divinejee.com/question/let-the-normal-at-a-point-p-on-the-curve-y2-3x2-y-10-0-intersect-the-yaxis-at-03-2-if-m"}]},{"@type":"QAPage","mainEntity":{"@type":"Question","name":"Question Let the normal at a point P on the curve y2 3x2 y 10 0 inter on Topic","text":"Let the normal at a point P on the curve\n
y² – 3x² + y + 10 = 0 intersect the y-axis at $\\left( {0,{3 \\over 2}} \\right)$\n. <...","answerCount":1,"upvoteCount":0,"datePublished":"2026-01-16","author":{"@type":"Organization","name":"DivineJEE"},"acceptedAnswer":{"@type":"Answer","text":"Check the detailed solution on the page.","upvoteCount":0,"url":"https://www.divinejee.com/question/let-the-normal-at-a-point-p-on-the-curve-y2-3x2-y-10-0-intersect-the-yaxis-at-03-2-if-m"}}}]}20:T412,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