
The locus of the mid-points of the chords of the circle \[{x^2} + {y^2} = 16\;\] which are tangents to the hyperbola \[9{x^2} - 16{y^2} = 144\;\] is :
A. \[{\left( {{x^2} + {y^2}} \right)^2} = 16{x^2} - 9{y^2}\]
B. \[{\left( {{x^2} + {y^2}} \right)^2} = 9{x^2} - 16{y^2}\]
C. \[{\left( {{x^2} - {y^2}} \right)^2} = 16{x^2} - 9{y^2}\]
D.None of these
Answer
555k+ views
Hint: To get the locus of mid points mid points of the chord of contact first suppose an unknown point as the mid point then just write the equation of chord of contact of circle then apply the condition of tangency on hyperbola.
Complete step-by-step answer:
Let \[P\left( {{x_1},{y_1}} \right)\;\] be the mid point of a chord of the circle.
The equation of the chord is \[T = {S_1}\]
On putting the coordinates of the mid point we get the equation of chord
\[
\Rightarrow x{x_1} + y{y_1} - 16 = {x_1}^2 + {y_1}^2 - 16 \\
y = - \dfrac{{{x_1}}}{{{y_1}}}x + \dfrac{{{x_1}{^2} + {y_{1}}^2}}{{{y_1}}} \\
\]
Since it chord touches the hyperbola \[\dfrac{{{x^2}}}{{{4^2}}} - \dfrac{{{y^2}}}{{{3^2}}} = 1\]
∴ \[{\left( {\dfrac{{{x_1}^{2} + {y_1}{^2}}}{{{y_1}}}} \right)^2} = {4^2}.\dfrac{{{x_{1}}^{2}}}{{{y_1}^2}} - {3^2}\;\;\;\;\;\;\;\;\;\;\] \[[\because {c^2} = {a^2}{m^2} - {b^2}] \]
∴ Locus of \[P\left( {{x_1},{y_1}} \right)\;\] is
\[
\dfrac{{{{\left( {{x^2} + {y^2}} \right)}^2}}}{{{y^2}}} = \dfrac{{16{x^2} - 9{y^2}}}{{{y^2}}} \\
\]
Or on cancelling the like terms
\[{\left( {{x^2} + {y^2}} \right)^2} = 16{x^2} - 9{y^2}\]
Hence the locus of mid point of the chord of the circle which is tangent to the hyperbola is
\[{\left( {{x^2} + {y^2}} \right)^2} = 16{x^2} - 9{y^2}\]
Therefore the option A is the correct answer for this question.
So, the correct answer is “Option A”.
Note: In this type of question where it is asked to find the locus of something, follow all the given conditions accordingly to the question then we will get the required equation of locus. Here we equate the equation of the midpoint of the chord of circle and tangent to the hyperbola to find the assumed variable.
Complete step-by-step answer:
Let \[P\left( {{x_1},{y_1}} \right)\;\] be the mid point of a chord of the circle.
The equation of the chord is \[T = {S_1}\]
On putting the coordinates of the mid point we get the equation of chord
\[
\Rightarrow x{x_1} + y{y_1} - 16 = {x_1}^2 + {y_1}^2 - 16 \\
y = - \dfrac{{{x_1}}}{{{y_1}}}x + \dfrac{{{x_1}{^2} + {y_{1}}^2}}{{{y_1}}} \\
\]
Since it chord touches the hyperbola \[\dfrac{{{x^2}}}{{{4^2}}} - \dfrac{{{y^2}}}{{{3^2}}} = 1\]
∴ \[{\left( {\dfrac{{{x_1}^{2} + {y_1}{^2}}}{{{y_1}}}} \right)^2} = {4^2}.\dfrac{{{x_{1}}^{2}}}{{{y_1}^2}} - {3^2}\;\;\;\;\;\;\;\;\;\;\] \[[\because {c^2} = {a^2}{m^2} - {b^2}] \]
∴ Locus of \[P\left( {{x_1},{y_1}} \right)\;\] is
\[
\dfrac{{{{\left( {{x^2} + {y^2}} \right)}^2}}}{{{y^2}}} = \dfrac{{16{x^2} - 9{y^2}}}{{{y^2}}} \\
\]
Or on cancelling the like terms
\[{\left( {{x^2} + {y^2}} \right)^2} = 16{x^2} - 9{y^2}\]
Hence the locus of mid point of the chord of the circle which is tangent to the hyperbola is
\[{\left( {{x^2} + {y^2}} \right)^2} = 16{x^2} - 9{y^2}\]
Therefore the option A is the correct answer for this question.
So, the correct answer is “Option A”.
Note: In this type of question where it is asked to find the locus of something, follow all the given conditions accordingly to the question then we will get the required equation of locus. Here we equate the equation of the midpoint of the chord of circle and tangent to the hyperbola to find the assumed variable.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

