Question

Find the equation of the tangent to the hyperbola \[4{x^2} - 9{y^2} = 1\]which is parallel to the line \[4y = 5x + 7\].

Answer 1

Hint: Here we convert the given equation of hyperbola into an equation such that we are easily able to compare it with the general equation of hyperbola and write the values of a and b. Compare the equation of line with the general equation of line and write the value of slope of line. Since, the slope of parallel lines are equal, using the slope of a given line and the formula for the line to be a tangent to the hyperbola we find an equation of tangent.
* A line \[y = mx + c\]is a tangent to the hyperbola \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\]if \[{a^2}{m^2} - {b^2} = {c^2}\]
* General equation of a hyperbola is \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\]
* General equation of the straight line is \[y = mx + c\].
* Two lines are parallel if their slopes are equal.

Complete step-by-step 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.
\[ \Rightarrow \dfrac{{4y}}{4} = \dfrac{{5x + 7}}{4}\]
Cancel the same factors from numerator and denominator.
\[ \Rightarrow y = \dfrac{5}{4}x + \dfrac{7}{4}\]
Compare the equation of line to the general equation of line i.e. \[y = mx + c\].
\[m = \dfrac{5}{4}\]
Slope of the line is \[m = \dfrac{5}{4}\] … (1)
Since, we know parallel lines have same slope, then slope of the tangent of hyperbola is equal to the slope of the line \[4y = 5x + 7\]
\[\therefore m = \dfrac{5}{4}\]is the slope of the tangent to the hyperbola.
Now we have equation of hyperbola as \[4{x^2} - 9{y^2} = 1\]
We transform this equation such that it is similar to the general equation of hyperbola \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\].
We can write the equation \[4{x^2} - 9{y^2} = 1\] as \[\dfrac{{{x^2}}}{{{{(\dfrac{1}{2})}^2}}} - \dfrac{{{y^2}}}{{{{(\dfrac{1}{3})}^2}}} = 1\]
On comparing with the general equation of hyperbola \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\], we get
\[a = \dfrac{1}{2},b = \dfrac{1}{3}\].
We know a line is a tangent to the hyperbola if \[{a^2}{m^2} - {b^2} = {c^2}\]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\[{a^2}{m^2} - {b^2} = {c^2}\].
\[ \Rightarrow {c^2} = {\left( {\dfrac{1}{2}} \right)^2}{\left( {\dfrac{5}{4}} \right)^2} - {\left( {\dfrac{1}{3}} \right)^2}\]
\[ \Rightarrow {c^2} = \dfrac{1}{4} \times \dfrac{{25}}{{16}} - \dfrac{1}{9}\]
\[ \Rightarrow {c^2} = \dfrac{{25}}{{64}} - \dfrac{1}{9}\]
Take LCM in RHS of the equation
\[ \Rightarrow {c^2} = \dfrac{{25 \times 9 - 64}}{{64 \times 9}}\]
\[ \Rightarrow {c^2} = \dfrac{{225 - 64}}{{16 \times 9 \times 4}}\]
Write the denominator in the form of a square of numbers.
\[ \Rightarrow {c^2} = \dfrac{{161}}{{{4^2} \times {3^2} \times {2^2}}}\]
Use the property \[{a^m}.{b^m} = {(ab)^m}\]
\[ \Rightarrow {c^2} = \dfrac{{161}}{{{{(4 \times 3 \times 2)}^2}}}\]
\[ \Rightarrow {c^2} = \dfrac{{161}}{{{{(24)}^2}}}\]
Take the square root on both sides of the equation.
\[ \Rightarrow \sqrt {{c^2}} = \sqrt {\dfrac{{161}}{{{{(24)}^2}}}} \]
Cancel square root with square power on both sides of the equation.
\[ \Rightarrow c = \dfrac{{\sqrt {161} }}{{24}}\]
Now we know the equation of tangent will be \[y = mx + c\].
Substitute the value of \[m = \dfrac{5}{4},c = \dfrac{{\sqrt {161} }}{{24}}\]in equation of tangent.
\[ \Rightarrow y = \dfrac{5}{4}x + \dfrac{{\sqrt {161} }}{{24}}\]

Therefore, equation of tangent to the hyperbola is \[y = \dfrac{5}{4}x + \dfrac{{\sqrt {161} }}{{24}}\].

Note: Students are likely to get confused while converting the given equation of hyperbola in a way so we can compare it to the general equation of hyperbola. Students will try to take LCM of 4 and 9 and divide both sides by the LCM to make an equation like hyperbola which is wrong because then we will not have RHS as 1. Focus on converting the LHS of the equation without making changes in the RHS.