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The slope of the tangent which touches both the parabolas $y^2$ = $4ax$ and the parabola $x^2=-32y$ how do we find the slope of common tangent if I assume the slope of one of the cords and I find the relation that would hold between the two or should start some other way round because I cannot understand how would this happen thanks in advance .

Nitro phenol
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Hint: A tangent to the parabola $y^2=4ax$ is of the form: $$y=mx+\frac{a}{m}$$ As this must also be the tangent to the second parabola $x^2=-32y$ , the discriminant of the quadratic equation formed when we put $y=mx+\frac{a}{m}$ in $x^2=-32y$ must be $0$.

Nikunj
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  • that is what I am unable to understand how come the slope of the tangent be mx+a/m form plz elaborate this @Nikunj – Nitro phenol Mar 29 '16 at 04:12
  • Do this: Take a general equation of a line as $y=mx+c$, Put it in the equation of a parabola $y^2=4ax$, you will get something like $(mx+c)^2=4ax$ as the tangent intersects the parabola only at one point, the discriminant of the quadratic must be 0. So try to make the discriminant $0$ – Nikunj Mar 29 '16 at 04:17
  • what would we obtain from there – Nitro phenol Mar 29 '16 at 04:21
  • You would obtain $c=\frac{a}{m}$ – Nikunj Mar 29 '16 at 04:24
  • okay then we satisfy the same equation in the second equation and get the value of m from there right ? – Nitro phenol Mar 29 '16 at 04:25
  • You would have to make the discriminant $0$ because the given line is tangent to both parabolas – Nikunj Mar 29 '16 at 04:26
  • discriminant in the new equitation obtained by substituting the value of y as obtained y= mx+a/m right – Nitro phenol Mar 29 '16 at 04:29
  • Yes, that's correct. – Nikunj Mar 29 '16 at 04:29
  • ok are there similar general equation for the tangents to ellipses also if yes please tell me because I am preparing for the JEE this year and am practising this topic right now – Nitro phenol Mar 29 '16 at 04:33