Finding the value of k when the normal to the curve is parallel to the given line

Asked Jun 09 '23 at 13:30

Active Jun 09 '23 at 14:52

Viewed 56 times

The normal to the curve $\ y=2x^2+kx-3$ at the point $(3, -6)$ is parallel to the line $x+5y=10$.

Find the value of $k$.
Find the coordinates of the point where the normal meets the curve again.

I am trying to find the value of $k$ but my answer is not the same as the displayed one. Firstly, I find the gradient of the curve and of the line and equal them but the answer is not right.

From my answers, the gradient of the curve is $12+k$ and the gradient of the line is $-1\over 5$. I made the two equations equal to find k but the answer is different. It said the answer for k is $-7$. Mine is $-61\over 5$.

edited Jun 09 '23 at 14:52

asked Jun 09 '23 at 13:30

Kyooo

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What is your answer and how did you get it? – Another User Jun 09 '23 at 13:31
Quick beginner guide for asking a well-received question + please avoid "no clue" questions – Anne Bauval Jun 09 '23 at 13:33
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Line is parallel to normal to the curve, then gradient of curve is not equal to gradient of the line – Ivan Kaznacheyeu Jun 09 '23 at 13:33
Are the gradients of the curve and the line not the same? – Kyooo Jun 09 '23 at 13:35
Please rephrase your title so that it is more descriptive of your question – Digitallis Jun 09 '23 at 14:34
@Kyooo The normal to the curve is perpendicular to the curve, by definition. As such, if the gradient of the normal to the curve is $m$, the gradient of the tangent to the curve is $-\frac{1}{m}$. – Lemmon Jun 09 '23 at 14:38
Ahhh I get it! It has to be $ {-1\over (12+k)} = {-1\over 5} $ instead of $(12+k) = {-1\over 5}$! – Kyooo Jun 09 '23 at 14:59
Thank you so much! I forget to do the reciprocal of the gradient value!!!! – Kyooo Jun 09 '23 at 15:00

Finding the value of k when the normal to the curve is parallel to the given line

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