Question

The equation of the
the tangent at vertex to the
parabola 4y^2+ 6x = 8y + 7 is
(1)x= 11/6
(2) y = 2
(3) x=-11/6

(4)y= -2​

Answer 1

Answer:

x=11/6

Step-by-step explanation:

First find the vertex if the parabola 4y²+6x=8y+7 ...... (1)

Rearranging the above equation we get, 4y²-8y+4=-6x+11

⇒$4(y-1)^{2}=-6(x-\frac{11}{6} )$

⇒$(y-1)^{2}=-\frac{6}{4}  (x-\frac{11}{6} )$

Hence, the vertex of the parabola is ($\frac{11}{6}, 1$).

Now, we have to find the slope of the tangent to the parabola at the vertex.

Differentiating equation (1), with respect yo x, we get

$8y\frac{dy}{dx} +6=8\frac{dy}{dx}$

⇒ $\frac{dy}{dx} =\frac{3}{4y-4}$

⇒$[\frac{dy}{dx}] _{(\frac{11}{6},1) } =$∞

So, the tangent is parallel to y axis and passing through the vertex.

Therefore, the equation of the tangent will be, x= 11/6. (Answer)

Answer 2

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