Would someone be so smart and so kind to find the Tangent lineS to the curve y=x^2 that pass through (0,-a^2), a>0 ????

Thanks in advance.

Consider the tangent to the parabola at (b, b^2). If this tangent is to pass through the point (0, -a^2) then the gradient of the tangent is . But the gradient of the tangent is also .

Equate the two expressions and solve for b (in terms of a).

Knowing the gradient of the tangent and a point lying on the tangent, you shuld be able to write down the equation of the tangent.

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you need two things:

(1) Equation of line in slope-point form(i.e. determining equaation with slope and one point)

(2) Formula to find slope of tangent

Do you know above two?

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Malay


raah pakad tu ek chala chal, paa jayega madhushala

I know them
but when it came to solving this question, I was a total failure.

Isn't the line tangent at x=0 of f(x)=x^2 horizontal on the x-axis. IE, y=0? I don't see ho there could be any other solution. Mr. Fantastic. I read your post. Could you ealaborate?

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well I know for sure that the line can not be tangent at x=0

Yeah, the codition a>0 says that, but , therefore 0. Mr. Fantastic understands better than I. I'm like you. I'd like to know what's going on here.

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Please go back and read what I posted (by the way I fixed a small typo). What part do you not understand? Where do you get stuck in following and applying what I suggested?

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But note that never touches the parabola! As mr fantastic said, consider another arbitrary point . Find the slope of the tangent at , and then use the point to construct the slope of the tangent line through that point and the equation of the tangent line (all in terms of ).

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Duh.

Yeah...For some reason, I went stupid for a minute. I'm sorry that I got involved here. I probably confused the poor guy even more.

Sorry everbody.

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It is ok
dont worry bud and I was not confused, I know the solution well now.