CUET Preparation Today
If the equation of the tangent to the curve $y^2=a x^3+b$ at the point (2, 3) is $y=4 x-5$, then |
$a=2, b=7$ $a=7, b=2$ $a=2, b=-7$ $a=-2, b=7$ |
$a=2, b=-7$ |
curve: $y^2=ax^3+b$ so at (2, 3) $⇒ 3^2=a×2^3+b$ so $8a+b=9$ ...(1) tangent: $y=4x-5$ $Slope]_{at(2,3)}=4$ $⇒y'=4$ So tangent of curve $2yy'=30x^2$ so $2×3×4=3×a×2^2$ $⇒a=2$ from (i) → $b=-7$ |
