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Question

Answers

A. $\dfrac{5}{4}$

B. $\dfrac{4}{5}$

C. $\dfrac{1}{2}$

D. None of these

Answer

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We are given a conic section with equation \[9{x^2}-16{y^2} + 18x + 32y-151 = 0\] and we need to find out the e which is the eccentricity of the curve or the conic.

We will start of by defining what we mean by e or eccentricity of a conic –

The eccentricity of a conic section is defined to be the distance from any point on the conic section to its focus, divided by the perpendicular distance from that point to the nearest directrix.

The value of e remains constant for a given throughout.

Since, in the cases where we have to solve questions on conic section the first approach should be simplifying it so that it comes in the general form of any conic section.

We will simplify the given conic by forming perfect squares as shown,

\[9{x^2}-16{y^2} + 18x + 32y-151 = 0\]

First, we need to segregate the terms of x and y. That is,

\[(9{x^2} + 18x) + (-16{y^2} + 32y)-151 = 0\]

We will start making perfect squares by making the coefficient of 2 degree terms as one. This is shown as –

\[9({x^2} + 2x)-16({y^2} - 2y)-151 = 0\]

Now for perfect we need to add and subtract 1 to each bracket.

\[9[({x^2} + 2x + 1) - 1]-16[({y^2} - 2y + 1) - 1]-151 = 0\]

\[9[{(x + 1)^2} - 1]-16[{(y - 1)^2} - 1]-151 = 0\]

Now we will put the constants aside and also divide by them to make 1 as the RHS.

\[9{(x + 1)^2} - 9-16{(y - 1)^2} + 16-151 = 0\]

Taking all the constants to RHS

\[9{(x + 1)^2}-16{(y - 1)^2} = 144\]

Dividing both sides by 144 we get,

\[\dfrac{{{{(x + 1)}^2}}}{{16}}-\dfrac{{{{(y - 1)}^2}}}{9} = 1\]

\[\dfrac{{{{(x + 1)}^2}}}{{{4^2}}}-\dfrac{{{{(y - 1)}^2}}}{{{3^2}}} = 1\]

Now from this equation we get the values of a and b which are 4 and 3 respectively. Since, it is a hyperbola we have the equation of e for it as:

\[e = \dfrac{{\sqrt {{a^2} + {b^2}} }}{a} = \dfrac{{\sqrt {{4^2} + {3^2}} }}{4} = \dfrac{5}{4}\]