Question

Find the equation of the tangent to the curve at the given point and show that the sum of its intercepts on its axes is constant

Answer 1

\[\Huge \sqrt{x}+\sqrt{y}=\sqrt{a}, point~(x_1,y_1)\]

Answer 2

Attempt:
\[\LARGE \frac{dy}{dx}=- \sqrt{\frac{y_0}{x_0}}\]
\[\LARGE (y-y_0)= - \sqrt{\frac{y_0}{x_0}}(x-x_0)\]
don't know if this equation is correct or not but my ans is wrong

Answer 3

Your solution is ok, maybe is wrong becuase you have used (xo, yo) instead of (x1,y1) as stated in the problem

Answer 4

A question of notation, I guess...

Answer 5

its x1,y1 only sorry and no i didn't even mean that..answer is completely different..

Answer 6

\[\LARGE \frac{x}{\sqrt{x_1}}+\frac{y}{\sqrt{y1}}=\sqrt{a}\]
is the answer..

Answer 7

I am not even getting "a" in my answer o_O i wonder how they are getting

Answer 8

@ganeshie8

Answer 9

but what about the tangent !

Answer 10

actually your equation is exact same as the answer
you just need to simplify @DLS

Answer 11

\(\large y-y_1 = -\sqrt{\frac{y_1}{x_1}}(x-x_1)\)

divide \(\sqrt{y_1}\) both sides

\(\large \frac{y}{\sqrt{y_1}}-\frac{y_1}{\sqrt{y_1}} = -\sqrt{\frac{1}{x_1}}(x-x_1)\)

\(\large \frac{y}{\sqrt{y_1}}-\frac{y_1}{\sqrt{y_1}} = -\frac{x}{\sqrt{x_1}} + \sqrt{x_1}\)

\(\large \frac{y}{\sqrt{y_1}} + \frac{x}{\sqrt{x_1}} = \frac{y_1}{\sqrt{y_1}} + \sqrt{x_1}\)


\(\large \frac{y}{\sqrt{y_1}} + \frac{x}{\sqrt{x_1}} = \sqrt{y_1} + \sqrt{x_1}\)

\(\large \frac{y}{\sqrt{y_1}} + \frac{x}{\sqrt{x_1}} = \sqrt{a}\)