
Let $a$ and $b$ be positive real numbers such that $a > 1$ and $b < a$ . Let $P$ be a point in the first quadrant that lies on the hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$. Suppose the tangent to the hyperbola at $P$ passes through the point $\left( {1,0} \right)$, and suppose the normal to the hyperbola at $P$ cuts off equal intercepts on the coordinate axes. Let $\Delta $ denote the area of the triangle formed by the tangent at $P$, the normal at $P$ and the x-axis. If e denotes the eccentricity of the hyperbola, then which of the following statements is/are TRUE?
A. $1 < e < \sqrt 2 $
B. $\sqrt 2 < e < 2$
C. $\Delta = {a^4}$
D. $\Delta = {b^4}$
Answer
163.2k+ views
Hint: First, consider the coordinates of the point $P$. Then use the coordinates of the point $P$ and the point $\left( {1,0} \right)$ to calculate the equation of the tangent. Then using the slope of the tangent find the values of $a$ and $b$ . Substitute the values of $a$ and $b$ in the eccentricity formula to calculate the eccentricity. In the end, apply the properties and formula of the area of an isosceles triangle to get the required answer.
Formula Used: The equation of a tangent to the hyperbola passing from the point $\left( {{x_1},{y_1}} \right)$ in parametric form is: $\dfrac{{x{x_1}}}{{{a^2}}} - \dfrac{{y{y_1}}}{{{b^2}}} = 1$
Eccentricity of a hyperbola: $e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} $
Area of a triangle: $\dfrac{1}{2} \times base \times height$
Slope formula: The slope of a line passing through the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$.
Complete step by step solution:
The given equation of a hyperbola is $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$, where $a > 1$ and $b < a$ .
Let $\left( {a\sec\left( u \right),b\tan\left( u \right)} \right)$ be the coordinates of the point $P$ that lies on the hyperbola.
Image: A hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ with tangent passing through the point $\left( {1,0} \right)$
The tangent is passes through the point $P\left( {a\sec\left( u \right),b\tan\left( u \right)} \right)$.
Apply the general formula of the tangent to the hyperbola in parametric form.
So, the equation of the tangent is,
$\dfrac{{x\left( {a\sec\left( u \right)} \right)}}{{{a^2}}} - \dfrac{{y\left( {b\tan\left( u \right)} \right)}}{{{b^2}}} = 1$
$ \Rightarrow \dfrac{{x\sec\left( u \right)}}{a} - \dfrac{{y\tan\left( u \right)}}{b} = 1$
Since the tangent is also passes through the point $A\left( {1,0} \right)$.
Substitute $\left( {1,0} \right)$ in above equation.
$\dfrac{{\left( 1 \right)\sec\left( u \right)}}{a} - \dfrac{{\left( 0 \right)\tan\left( u \right)}}{b} = 1$
$ \Rightarrow \dfrac{{\sec\left( u \right)}}{a} = 1$
$ \Rightarrow \sec\left( u \right) = a$
Slope of the tangent is 1.
Apply the slope formula for the tangent.
$\dfrac{{b\tan\left( u \right) - 0}}{{a\sec\left( u \right) - 1}} = 1$
Simplify the above equation.
$\dfrac{{b\tan\left( u \right) - 0}}{{\left( {\sec\left( u \right)} \right)\sec\left( u \right) - 1}} = 1$ [since $a = \sec\left( u \right)$]
$ \Rightarrow \dfrac{{b\tan\left( u \right)}}{{\sec^{2}\left( u \right) - 1}} = 1$
$ \Rightarrow \dfrac{{b\tan\left( u \right)}}{{\tan^{2}\left( u \right)}} = 1$ [Since $\sec^{2}\left( u \right) - 1 = \tan^{2}\left( u \right)$]
$ \Rightarrow \dfrac{b}{{\tan\left( u \right)}} = 1$
$ \Rightarrow b = \tan\left( u \right)$
Substitute $a = \sec\left( u \right)$ and $b = \tan\left( u \right)$ in the formula of eccentricity.
$e = \sqrt {1 + \dfrac{{\tan^{2}\left( u \right)}}{{\sec^{2}\left( u \right)}}} $
$ \Rightarrow e = \sqrt {1 + \dfrac{{\left( {\dfrac{{\sin^{2}\left( u \right)}}{{\cos^{2}\left( u \right)}}} \right)}}{{\left( {\dfrac{1}{{\cos^{2}\left( u \right)}}} \right)}}} $ [since $\dfrac{1}{{\cos x}} = \sec x$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$]
$ \Rightarrow e = \sqrt {1 + \sin^{2}\left( u \right)} $
The range of $\sin^{2}\left( u \right)$ in first quadrant is $\left[ {0,1} \right]$.
So, the minimum value of $e$ in the first quadrant is,
$e = \sqrt {1 + \sin^{2}\left( 0 \right)} $
$ \Rightarrow e = \sqrt {1 + 0} $ [Since, $\sin\left( 0 \right) = 0$]
$ \Rightarrow e = 1$
And the maximum value of $e$ in the first quadrant is,
$e = \sqrt {1 + \sin^{2}\left( {\dfrac{\pi }{2}} \right)} $
$ \Rightarrow e = \sqrt {1 + 1} $ [Since $\sin\left( {\dfrac{\pi }{2}} \right) = 1$]
$ \Rightarrow e = \sqrt 2 $
Therefore, the eccentricity of the hyperbola lies between
$1 < e < \sqrt 2 $
Since, the vertices of the point $P$ are $\left( {\sec^{2}\left( u \right),\tan^{2}\left( u \right)} \right)$.
Apply the distance formula to calculate the distance between $A\left( {1,0} \right)$ and $P\left( {\sec^{2}\left( u \right),\tan^{2}\left( u \right)} \right)$.
$AP = \sqrt {{{\left( {\sec^{2}\left( u \right) - 1} \right)}^2} + {{\left( {\tan^{2}\left( u \right)} \right)}^2}} $
$ \Rightarrow AP = \sqrt {{{\left( {{{\tan }^2}\left( u \right)} \right)}^2} + {{\left( {\tan^{2}\left( u \right)} \right)}^2}} $
$ \Rightarrow AP = \sqrt {2\tan^{4}\left( u \right)} $
$ \Rightarrow AP = \sqrt 2 \tan^{2}\left( u \right)$
Since the $\Delta APB$ is an isosceles triangle.
So, $AP = BP$
Apply the formula of the area of a triangle.
$\Delta = \dfrac{1}{2} \times AP \times BP$
$ \Rightarrow \Delta = \dfrac{1}{2} \times {\left( {AP} \right)^2}$ [since, $AP = BP$]
$ \Rightarrow \Delta = \dfrac{1}{2} \times {\left( {\sqrt 2 \tan^{2}\left( u \right)} \right)^2}$
$ \Rightarrow \Delta = \dfrac{1}{2} \times \left( {2\tan^{4}\left( u \right)} \right)$
$ \Rightarrow \Delta = \tan^{4}\left( u \right)$
$ \Rightarrow \Delta = {b^4}$
Option ‘A’ and ‘D; are correct
Note: The equation of tangents to the hyperbola at the point is: $\left( {{x_1},{y_1}} \right)$ is $\dfrac{{x{x_1}}}{{{a^2}}} - \dfrac{{y{y_1}}}{{{b^2}}} = 1$.
Two tangents can be drawn to the hyperbola from an external point: $\left( {{x_1},{y_1}} \right)$ to the hyperbola.
The equation of the normal to hyperbola at the point is: $\left( {{x_1},{y_1}} \right)$ is $\dfrac{{{a^2}x}}{{{x_1}}} - \dfrac{{{b^2}y}}{{{y_1}}} = 1$.
Formula Used: The equation of a tangent to the hyperbola passing from the point $\left( {{x_1},{y_1}} \right)$ in parametric form is: $\dfrac{{x{x_1}}}{{{a^2}}} - \dfrac{{y{y_1}}}{{{b^2}}} = 1$
Eccentricity of a hyperbola: $e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} $
Area of a triangle: $\dfrac{1}{2} \times base \times height$
Slope formula: The slope of a line passing through the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$.
Complete step by step solution:
The given equation of a hyperbola is $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$, where $a > 1$ and $b < a$ .
Let $\left( {a\sec\left( u \right),b\tan\left( u \right)} \right)$ be the coordinates of the point $P$ that lies on the hyperbola.
Image: A hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ with tangent passing through the point $\left( {1,0} \right)$
The tangent is passes through the point $P\left( {a\sec\left( u \right),b\tan\left( u \right)} \right)$.
Apply the general formula of the tangent to the hyperbola in parametric form.
So, the equation of the tangent is,
$\dfrac{{x\left( {a\sec\left( u \right)} \right)}}{{{a^2}}} - \dfrac{{y\left( {b\tan\left( u \right)} \right)}}{{{b^2}}} = 1$
$ \Rightarrow \dfrac{{x\sec\left( u \right)}}{a} - \dfrac{{y\tan\left( u \right)}}{b} = 1$
Since the tangent is also passes through the point $A\left( {1,0} \right)$.
Substitute $\left( {1,0} \right)$ in above equation.
$\dfrac{{\left( 1 \right)\sec\left( u \right)}}{a} - \dfrac{{\left( 0 \right)\tan\left( u \right)}}{b} = 1$
$ \Rightarrow \dfrac{{\sec\left( u \right)}}{a} = 1$
$ \Rightarrow \sec\left( u \right) = a$
Slope of the tangent is 1.
Apply the slope formula for the tangent.
$\dfrac{{b\tan\left( u \right) - 0}}{{a\sec\left( u \right) - 1}} = 1$
Simplify the above equation.
$\dfrac{{b\tan\left( u \right) - 0}}{{\left( {\sec\left( u \right)} \right)\sec\left( u \right) - 1}} = 1$ [since $a = \sec\left( u \right)$]
$ \Rightarrow \dfrac{{b\tan\left( u \right)}}{{\sec^{2}\left( u \right) - 1}} = 1$
$ \Rightarrow \dfrac{{b\tan\left( u \right)}}{{\tan^{2}\left( u \right)}} = 1$ [Since $\sec^{2}\left( u \right) - 1 = \tan^{2}\left( u \right)$]
$ \Rightarrow \dfrac{b}{{\tan\left( u \right)}} = 1$
$ \Rightarrow b = \tan\left( u \right)$
Substitute $a = \sec\left( u \right)$ and $b = \tan\left( u \right)$ in the formula of eccentricity.
$e = \sqrt {1 + \dfrac{{\tan^{2}\left( u \right)}}{{\sec^{2}\left( u \right)}}} $
$ \Rightarrow e = \sqrt {1 + \dfrac{{\left( {\dfrac{{\sin^{2}\left( u \right)}}{{\cos^{2}\left( u \right)}}} \right)}}{{\left( {\dfrac{1}{{\cos^{2}\left( u \right)}}} \right)}}} $ [since $\dfrac{1}{{\cos x}} = \sec x$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$]
$ \Rightarrow e = \sqrt {1 + \sin^{2}\left( u \right)} $
The range of $\sin^{2}\left( u \right)$ in first quadrant is $\left[ {0,1} \right]$.
So, the minimum value of $e$ in the first quadrant is,
$e = \sqrt {1 + \sin^{2}\left( 0 \right)} $
$ \Rightarrow e = \sqrt {1 + 0} $ [Since, $\sin\left( 0 \right) = 0$]
$ \Rightarrow e = 1$
And the maximum value of $e$ in the first quadrant is,
$e = \sqrt {1 + \sin^{2}\left( {\dfrac{\pi }{2}} \right)} $
$ \Rightarrow e = \sqrt {1 + 1} $ [Since $\sin\left( {\dfrac{\pi }{2}} \right) = 1$]
$ \Rightarrow e = \sqrt 2 $
Therefore, the eccentricity of the hyperbola lies between
$1 < e < \sqrt 2 $
Since, the vertices of the point $P$ are $\left( {\sec^{2}\left( u \right),\tan^{2}\left( u \right)} \right)$.
Apply the distance formula to calculate the distance between $A\left( {1,0} \right)$ and $P\left( {\sec^{2}\left( u \right),\tan^{2}\left( u \right)} \right)$.
$AP = \sqrt {{{\left( {\sec^{2}\left( u \right) - 1} \right)}^2} + {{\left( {\tan^{2}\left( u \right)} \right)}^2}} $
$ \Rightarrow AP = \sqrt {{{\left( {{{\tan }^2}\left( u \right)} \right)}^2} + {{\left( {\tan^{2}\left( u \right)} \right)}^2}} $
$ \Rightarrow AP = \sqrt {2\tan^{4}\left( u \right)} $
$ \Rightarrow AP = \sqrt 2 \tan^{2}\left( u \right)$
Since the $\Delta APB$ is an isosceles triangle.
So, $AP = BP$
Apply the formula of the area of a triangle.
$\Delta = \dfrac{1}{2} \times AP \times BP$
$ \Rightarrow \Delta = \dfrac{1}{2} \times {\left( {AP} \right)^2}$ [since, $AP = BP$]
$ \Rightarrow \Delta = \dfrac{1}{2} \times {\left( {\sqrt 2 \tan^{2}\left( u \right)} \right)^2}$
$ \Rightarrow \Delta = \dfrac{1}{2} \times \left( {2\tan^{4}\left( u \right)} \right)$
$ \Rightarrow \Delta = \tan^{4}\left( u \right)$
$ \Rightarrow \Delta = {b^4}$
Option ‘A’ and ‘D; are correct
Note: The equation of tangents to the hyperbola at the point is: $\left( {{x_1},{y_1}} \right)$ is $\dfrac{{x{x_1}}}{{{a^2}}} - \dfrac{{y{y_1}}}{{{b^2}}} = 1$.
Two tangents can be drawn to the hyperbola from an external point: $\left( {{x_1},{y_1}} \right)$ to the hyperbola.
The equation of the normal to hyperbola at the point is: $\left( {{x_1},{y_1}} \right)$ is $\dfrac{{{a^2}x}}{{{x_1}}} - \dfrac{{{b^2}y}}{{{y_1}}} = 1$.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Instantaneous Velocity - Formula based Examples for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

JEE Advanced 2025 Notes

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Total MBBS Seats in India 2025: Government and Private Medical Colleges

NEET Total Marks 2025
