If P(x₁, y₁), Q(x₂, y₂), R(x₃, y₃) and S(x₄, y₄) are 4 concyclic points on rectangul

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10-2. Parabola, Ellipse, Hyperbola

normal

If $ P(x_1, y_1), Q(x_2, y_2), R(x_3, y_3) $ and $ S(x_4, y_4) $ are $4 $ concyclic points on the rectangular hyperbola $x y = c^2$ , the co-ordinates of the orthocentre of the triangle $ PQR$ are :

$(x_4, - y_4)$

$(x_4, y_4)$

$(- x_4, - y_4)$

$(- x_4, y_4)$

Solution

A rectangular hyperbola circumscribing a $ \Delta$ also passes through its orthocentre if $\left( {c{t_i},\,\frac{c}{{{t_i}}}} \right)$ where $i = 1, 2, 3 $ are the vertices of the $ \Delta $ then therefore orthocentre is $\left( {\frac{{ – c}}{{{t_1}{t_2}{t_3}}}, – c{t_1}{t_2}{t_3}} \right)$ , where $t_1 t_2 t_3 t_4 = 1.$ Hence orthocentre is $\left( { – c{t_4},\,\,\frac{{ – c}}{{{t_4}}}} \right)$ = $(- x_4 , – y_4)$

Standard 11

Mathematics

The foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and the hyperbola $\frac{{{x^2}}}{{144}} – \frac{{{y^2}}}{{81}} = \frac{1}{{25}}$ coincide. Then the value of $b^2$ is

normal

A tangent to a hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$ intercepts a length of unity from each of the co-ordinate axes, then the point $(a, b)$ lies on the rectangular hyperbola

medium

The equation of the hyperbola whose foci are $(-2, 0)$ and $(2, 0)$ and eccentricity is $2$ is given by :-

normal

Consider a hyperbola $H : x ^{2}-2 y ^{2}=4$. Let the tangent at a point $P (4, \sqrt{6})$ meet the $x$ -axis at $Q$ and latus rectum at $R \left( x _{1}, y _{1}\right), x _{1}>0 .$ If $F$ is a focus of $H$ which is nearer to the point $P$, then the area of $\Delta QFR$ is equal to ……. .

hard

(JEE MAIN-2021)

If the vertices of a hyperbola be at $(-2, 0)$ and $(2, 0)$ and one of its foci be at $(-3, 0)$, then which one of the following points does not lie on this hyperbola?

hard

(JEE MAIN-2019)

If $ P(x_1, y_1), Q(x_2, y_2), R(x_3, y_3) $ and $ S(x_4, y_4) $ are $4 $ concyclic points on the rectangular hyperbola $x y = c^2$ , the co-ordinates of the orthocentre of the triangle $ PQR$ are :

Solution

Similar Questions

The foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and the hyperbola $\frac{{{x^2}}}{{144}} – \frac{{{y^2}}}{{81}} = \frac{1}{{25}}$ coincide. Then the value of $b^2$ is

A tangent to a hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$ intercepts a length of unity from each of the co-ordinate axes, then the point $(a, b)$ lies on the rectangular hyperbola

The equation of the hyperbola whose foci are $(-2, 0)$ and $(2, 0)$ and eccentricity is $2$ is given by :-

Consider a hyperbola $H : x ^{2}-2 y ^{2}=4$. Let the tangent at a point $P (4, \sqrt{6})$ meet the $x$ -axis at $Q$ and latus rectum at $R \left( x _{1}, y _{1}\right), x _{1}>0 .$ If $F$ is a focus of $H$ which is nearer to the point $P$, then the area of $\Delta QFR$ is equal to ……. .

If the vertices of a hyperbola be at $(-2, 0)$ and $(2, 0)$ and one of its foci be at $(-3, 0)$, then which one of the following points does not lie on this hyperbola?

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