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If the equation of the tangent to the curve `y^2=a x^3+b`at point `(2,3)i sy=4x-5`, then find the values of `aa n db`.A. `a=2, b=7`B. `a=7, b=2`C. `a=2, b= -7`D. `a= -2,b=7` |
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Answer» Correct Answer - C The equation of the curve is `y^(2)=ax^(3)+b.` `therefore 2y (dy)/(dx)=3ax^(2) rArr (dy)/(dx) = (3ax^(2))/(2y) rArr ((dy)/(dx))_((2","3))=2a` The equation of the tangent at (2, 3) is `y-3=2a(x-3) or, 2ax-y-6a+3=0` But, the equation of the tangent at (2, 3) is `y=4x-5.` Comparing these two equations, we get `2a=4 and -6a+3= -5 rArr a=2` As (2, 3) lies on the curve `y^(2)=ax^(3)+b.` `therefore 9=8a+b rArr 9=16+b rArr b = -7` |
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