Tag: normal to a hyperaboal

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Questions Related to normal to a hyperaboal

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The equation of the curve which is such that the protion of the axis of x cut off between the origin and tangent at any point is proportional to the ordinate of that point is _______________.

  1. $\log x = b y ^ { 2 } + a$
  2. $x = y ( a + b \log y )$
  3. $x = y ( b - a \log y )$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The problem describes a differential equation where the x-intercept of the tangent is proportional to the ordinate y. Solving this leads to the curve x = y(b - a log y).

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$, normals are drawn to curve $\left( {{{\left( {\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}}} \right)}^2} - 1} \right)\left( {\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}}} \right) = 0$.
Find the sum;  of abscissa of foot of all such normals.

  1. $\frac{{6{a^2}h}}{{\left( {{a^2} + {b^2}} \right)}}$
  2. $\frac{{8{a^2}h}}{{\left( {{a^2} + {b^2}} \right)}}$
  3. $\frac{{6a{h^2}}}{{\left( {{a^2} + {b^2}} \right)}}$
  4. $\frac{{8a{h^2}}}{{\left( {{a^2} + {b^2}} \right)}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is a complex geometry problem involving the sum of abscissae for normals to a hyperbola. The derived result for the sum is 8*a^2*h / (a^2 + b^2).

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

If the straight line $(a - 2) x - by + 4 = 0$ is normal to the hyperbola $xy = 1$ then which of the followings does not hold?

  1. $a > 1, b > 0$
  2. $a > 1, b < 0$
  3. $a < 1, b < 0$
  4. $a < 1, b > 0$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Every normal to $xy = 1$ must have positive slope as $\dfrac {-dx}{dy} = x^{2}$. So $\dfrac {a - 1}{b} > 0$.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The normal to the hyperbola $4x^2-9y^2=36$ meets the axes in $M$ and $N$ and the lines $MP$, $NP$ are drawn right angles at the axes. The locus of $P$ is the hyperbola 

  1. $9x^2-4y^2=169$
  2. $4x^2-9y^2=169$
  3. $3x^2-4y^2=169$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac {x^2}9-\dfrac {y^2}4=1$.Let $P(x _1, y _1)$ be the point on hyperbola.

Eqn of the normal is$\dfrac {a^2x}{x _1}-\dfrac {b^2y}{y _1}=a^2b^2\M=\left( \dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 } }  \right) { x } _{ 1 }=x\N=\left( \dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 } }  \right) { y } _{ 1 }=y\P=(x, y)$$x _1=\dfrac {a^2(x)}{a^2-b^2}$ $(x _1, y _1)$ lies at hyperbola.$y _1=\dfrac {b^2(y)}{a^2-b^2}$ $(x _1, y _1)$ lies at hyperbola.Now, $a^2=9, b^2=4$Therefore, $x _1=\dfrac {9x}5, y _1=\dfrac {4y}5$$\dfrac {x _1^2}9-\dfrac {y _1^2}4=1\\left(\dfrac {9x}5\right)^2\dfrac {x _1^2}9-\left(\dfrac {4y}5\right)^2\dfrac {y _1^2}4=1\\implies 9x^2-4y^2=25$

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

A normal to the hyperbola, $4x^2-9y^2=36$ meets the co-ordinate axes x and y at A and B, respectively. If the parallelogram $OABP$($O$ being the origin) is formed, then the locus of $P$ is?

  1. $4x^{2}+9y^{2}=121$
  2. $9x^{2}+4y^{2}=169$
  3. $4x^{2}-9y^{2}=121$
  4. $9x^{2}-4y^{2}=169$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For the hyperbola 4x^2 - 9y^2 = 36, the normal equation and the locus of the vertex P of the parallelogram OABP result in 9x^2 - 4y^2 = 169.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

Equation of the normal to the hyperbola $3x^2-y^2=3$ at $(2, -3)$ is?

  1. $x-2y-8=0$
  2. $3x-2y-12=0$
  3. $x+2y+4=0$
  4. $3x+2y-14=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the hyperbola 3x^2 - y^2 = 3, the derivative at (2, -3) gives the slope of the tangent. The normal slope is the negative reciprocal. Using the point-slope form, the equation is x - 2y - 8 = 0.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

Line x cos$\alpha $+yin$\alpha $=p is a normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $, if

  1. $a^{2}sec^{2}\alpha -b^{2}cosec^{2}\alpha =\frac{(a^{2}+b^{2})^{2}}{p^{^{2}}}$
  2. $a^{2}sec^{2}\alpha+b^{2}cosec^{2}\alpha =\frac{(a^{2}+b^{2})^{2}}{p^{^{2}}}$
  3. $a^{2}cos^{2}\alpha -b^{2}sin^{2}\alpha =\frac{(a^{2}+b^{2})^{2}}{p^{^{2}}}$
  4. $a^{2}cos^{2}\alpha+b^{2}sin^{2}\alpha =\frac{(a^{2}+b^{2})^{2}}{p^{^{2}}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The condition for the line x cos(alpha) + y sin(alpha) = p to be a normal to the hyperbola x^2/a^2 - y^2/b^2 = 1 is a^2 sec^2(alpha) - b^2 csc^2(alpha) = (a^2 + b^2)^2 / p^2.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

Line $ x \cos \alpha + y \sin \alpha = p $ is a normal to the hyperbola $ \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 $, if 

  1. $ a ^ { 2 } \sec ^ { 2 } \alpha - b ^ { 2 } \csc ^ { 2 } \alpha = \frac { \left( a ^ { 2 } + b ^ { 2 } \right) ^ { 2 } } { p ^ { 2 } } $
  2. $ a ^ { 2 } \sec ^ { 2 } x + b ^ { 2 } \csc ^ { 2 } \alpha = \frac { \left( a ^ { 2 } + b ^ { 2 } \right) ^ { 2 } } { p ^ { 2 } } $
  3. $ a ^ { 2 } \cos ^ { 2 } \alpha - b ^ { 2 } \sin ^ { 2 } \alpha = \frac { \left( a ^ { 2 } + b ^ { 2 } \right) ^ { 2 } } { p ^ { 2 } } $
  4. $ a ^ { 2 } \cos ^ { 2 } \alpha + b ^ { 2 } \sin ^ { 2 } \alpha = \frac { \left( a ^ { 2 } + b ^ { 2 } \right) ^ { 2 } } { p ^ { 2 } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is identical to the previous question (424646). The condition for the line to be a normal is a^2 sec^2(alpha) - b^2 csc^2(alpha) = (a^2 + b^2)^2 / p^2.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

A straight line is drawn parallel to the conjugate axis of the hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ to meet it and the conjugate hyperbola respectively in the point $P$ and $Q$. The normals at $p$ and $Q$ to the curves meet on 

  1. $x-axis$
  2. $y-axis$
  3. $y=x$
  4. $y=-x$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Equation of hyperbola $\to \dfrac {x^2}{9^2}-\dfrac {y^2}{6^2}=1$
Conjugate hyperbola $\to \dfrac {y^2}{6^2}-\dfrac {x^2}{9^2}=1$
Line $||$ to conjugate axis of hyperbola $(y-axis)$ is drawn to meet conjugate axis at $P$ & $Q$ which are symmetric points about $x-$axis then
$P(a\tan \theta , b\sin theta)$
$Q(a \tan \theta , -b\sin \theta)$
Normal at $P\to \dfrac {ax}{\tan \theta} +\dfrac {6y}{\sec \theta}=a^2+b^2$
Normal at $Q\to \dfrac {ax}{\tan \theta}-\dfrac {6y}{\sec \theta}=a^2+b^2$
Let us find out interrection
$\dfrac {b}{\sec \theta}=\dfrac {-6y}{\sec \theta} $
$y=0$
$9+$ lies on $x-$axis
$A$ is correct


Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

If the normal at $\left (ct _1,\dfrac { c}{t _1}\right)$ on the hyperbola $xy = c^2$ cuts the hyperbola again at $\left (ct _2, \dfrac {c}{t _2}\right)$, then $t _2^3 t _2$ $=$ 

  1. $2$
  2. $-2$
  3. $-1$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation of hyperbola is $xy=c^2$ and point $(ct _1,\dfrac{c}{t _1})$ lies on it.


Let us find the equation of the normal.


Equation of hyprbola can be written as $y=\dfrac{c^2}{x}$ and therefore slope of tangent is given by first derivative i.e. $\dfrac{dy}{dx}=−\dfrac{c^2}{x^2}$


hence slope of normal is given by $\dfrac{x^2}{c^2}$ and at $(ct _1,\dfrac{c}{t _1})$ is $t^2$ and its equation is


$y=t _1^2(x−ct _1)+\dfrac{c}{t _1}$


or $xt _1^3−yt _1−ct _1^4+c=0$


As this passes through $(ct _2,\dfrac{c}{t _2})$


$ct _2t _1^3−\dfrac{c}{t _2}t _1−ct _1^4+c=0$


or $ct _1^3(t _2−t _1)+\dfrac{c}{t _2}(t _2−t _1)=0$


as $t _1\neq t _2, t _1−t _2\neq 0$ and dividing by it we get


$ct _2^3=−\dfrac{c}{t _2}$


Or $ t _2^3t _2=−1$