[[1]] is the equation of the tangent to the hyperbola 4x2−9y2=1 which is parallel to the line 4y=5x+7.

Question

[[1]] is the equation of the tangent to the hyperbola 4x2−9y2=1  which is parallel to the line 4y=5x+7.

Accepted Answer

We are given the equation of line as 4y=5x+7
We find the value of the slope of the line.
Divide the equation of line by 4.

\Rightarrow \frac{4 y}{4} = \frac{5 x + 7}{4}

Cancel the same factors from numerator and denominator.

\Rightarrow y = \frac{5}{4} x + \frac{7}{4}

Compare the equation of line to the general equation of line i.e.

y = mx + c .

m = \frac{5}{4}

Slope of the line is

m = \frac{5}{4} \dots \dots \dots . (1)

Since, we know parallel lines have same slope, then slope of the tangent of hyperbola is equal to the slope of the line

4 y = 5 x + 7

∴ m = \frac{5}{4}

is the slope of the tangent to the hyperbola.
Now we have equation of hyperbola as

46 x^{2} - 9 y^{2} = 1

We transform this equation such that it is similar to the general equation of hyperbola

x^{2} a^{2} - y^{2} b^{2} = 1

We can write the equation

4 x^{2} - 9 y^{2} = 1

as

x^{2} {(12)}^{2} - y^{2} {(13)}^{2} = 1

On comparing with the general equation of hyperbola

x^{2} a^{2} - y^{2} b^{2} = 1

, we get a=12,b=13
We know a line is a tangent to the hyperbola if a2m2−b2=c2
where a, b are constants from hyperbola and m is the slope of the line.
Substituting the values of a, b and m in the equation a2m2−b2=c2
⇒c2=(12)2(54)2−(13)2

\Rightarrow c^{2} = \frac{1}{4} \times \frac{25}{16} - \frac{1}{9}

\Rightarrow c^{2} = \frac{25}{64} - \frac{1}{9}

Take LCM in RHS of the equation

\Rightarrow c^{2} = \frac{25 \times 9 - 64}{64 \times 9}

\Rightarrow c^{2} = \frac{225 - 64}{16 \times 9 \times 4}

Write the denominator in the form of a square of numbers.

\Rightarrow c^{2} = \frac{161}{4^{2} \times 3^{2} \times 2^{2}}

\Rightarrow c^{2} = \frac{161}{{(4 \times 3 \times 2)}^{2}}

\Rightarrow c^{2} = \frac{161}{{(24)}^{2}}

Take the square root on both sides of the equation.

\Rightarrow c^{2} = \frac{161}{{(24)}^{2}}

Cancel square root with square power on both sides of the equation.

\Rightarrow c = \frac{\sqrt{161}}{24}

Now we know the equation of tangent will be

y = mx + c .

Substitute the value of m=54,

c = \frac{\sqrt{161}}{24}

in the equation of tangent.

y = \frac{5}{4} x + \frac{\sqrt{161}}{24}

Therefore, equation of tangent to the hyperbola is

y = \frac{5}{4} x + \frac{\sqrt{161}}{24}

.

[[1]] is the equation of the tangent to the hyperbola 4x2−9y2=1 which is parallel to the line 4y=5x+7.

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