The slope of the curve `2y^2=a x^2+ba t(1,-1)` is `-1` Find `a , b`

Answer

If `a=2` then `b=0`, and

If `a=-2` then `b=4`.

Explanation

Given, the slope of the curve `2 y^{2}=a x^{2}+b` at `(1,-1)` is `-1`.

As, `(1,-1)` is on the given curve, then

`a+b=2` … eq (1)

Now, slope of the given curve at `(1,-1)` would be :


`\frac{d y}{d x}|_{(1,-1)}=.\frac{a x}{2 \sqrt{\frac{a x^{2}+b}{2}}}|_{(1,-1)}=\frac{a}{2 \sqrt{\frac{a+b}{2}}}`.


Given, `\frac{a}{2 \sqrt{\frac{a+b}{2}}}=-1`.

or, `a^{2}=4(\frac{a+b}{2})`

or, `a^{2}=2(a+b)`

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