Tangents are drawn to the hyperbola 4x2-y2 36
WebGraph 4x^2+9y^2=36 4x2 + 9y2 = 36 4 x 2 + 9 y 2 = 36 Find the standard form of the ellipse. Tap for more steps... x2 9 + y2 4 = 1 x 2 9 + y 2 4 = 1 This is the form of an ellipse. Use this form to determine the values used to find the center along with the major and minor axis of … WebSep 14, 2024 · Solution For (5) Tangents are drawn to the hyperbola 4x2−y2=36 at the point P and Q If these tangents intersect at the point T(0,3) then the area (in sq. units) of PTQ is …
Tangents are drawn to the hyperbola 4x2-y2 36
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WebOct 7, 2024 · Exp. (a) Tangents are drawn to the hyperbola 4x2−y2 =36 at the point P and Q . Tangent intersects at point T (0,3) Clearly, PQ is chord of contact. ∴ Equation of PQ is … WebJun 9, 2024 · Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
WebAnswer (1 of 7): Hyperbola: 4x²-y²=36 or x²/9 -y²/36 = 1. It’s tangent may be written as: ax/9 -by/36 = 1 and it is given that it passes through (0,3). Then, 0 - 3b/36=1 or b=- 12. WebNo views 1 minute ago Tangents are drawn to the hyperbola \ ( 4 x^ {2}-y^ {2}=36 \) at \ ( \mathrm {P} \) the points \ ( P \) and \ ( Q \). If these tangents intersect at the Illustrated...
WebTangents are drawn to the hyperbola \ ( 4 x^ {2}-y^ {2}=36 \) at the point \ ( \mathrm {P} \) and \ ( \mathrm {Q} \). If these tangents intersect at the point \ ( \mathrm {T} (0,3) \)... WebOct 15, 2024 · Tangents are drawn to the hyperbola 4x2 - y2 = 36 at the points P and Q. If these tangents intersect at the point T (0, 3), then the area (in sq units) of ΔPTQ is (a) …
WebOct 15, 2024 · Tangents are drawn to the hyperbola 4x2 - y2 = 36 at the points P and Q. If these tangents intersect at the ... ) 45√5 (b) 54√3 (c) 60√3 (d) 36√5. ... Tangents are drawn to the hyperbola `4x^2-y^2=36` at the points P and Q. If these tangents intersect at the point T(0,3) then the area (in sq units) o ...
WebTangents are drawn to the hyperbola 4x 2 - y 2 = 36 at the points P and Q. If these tangents intersect at the point T (0, 3) then the area (in sq. units) of Δ PTQ is : A 36 5 B 45 5 C 54 3 D 60 3 Check Answer 2 JEE Main 2024 (Offline) MCQ (Single Correct Answer) + 4 - 1 supreme face shirtWebThe correct options are. A x+y+3√3 = 0. C x+y−3√3 = 0. Let m be the slope of the tangent. As the tangent is perpendicular to the line x−y+4 = 0. So, m × 1 = -1 m = -1. Since, x2−4y2 =36 … supreme fanny packsWebFind the general equations of the lines tangent to the hyperbola 4x2-y2=36 and perpendicular to the line 5x-2y-4=0 This problem has been solved! You'll get a detailed … supreme fanny pack menWebTangents are drawn to the hyperbola 4x2−y2 =36 at the points P and Q. If these tangents intersect at the point T (0,3), then the area (in sq.units) of ΔP T Q is A 36√5 B 45√5 C 54√3 … supreme fiber dining tableWebSolution: Clearly PQ is a chord of contact, i.e., equation of PQ is T ≡ 0 ⇒ y = −12 Solving with the curve, 4x2 −y2 = 36 ⇒ x = ±3 5,y = −12 i.e., P (3 5,−12);Q(−3 5,−12);T (0,3) Area of … supreme field gear lightweight balaclavaWebNov 22, 2024 · Correct Answer - D We have, 4x2 − 9y2 = 36 ⇒ 8x − 18y dy dx = 0 ⇒ dy dx = 4x 9y 4 x 2 - 9 y 2 = 36 ⇒ 8 x - 18 y d y d x = 0 ⇒ d y d x = 4 x 9 y ∴ Slope of the tangent = 4x 9y ∴ Slope of the tangent = 4 x 9 y For this tangent to be perpendicular to the straight line 5x + 2y − 10 = 0, 5 x + 2 y - 10 = 0, we must have supreme field pullover redditWebMar 31, 2024 · Here, we have to find the equation of tangents to the given hyperbola such that the required tangent is perpendicular to the line x + 3y = 2 The given equation of hyperbola can be re-written as: x 2 1 − y 2 3 = 1 Let the slope of the line x + 3y = 2 be m 1 So, by comparing the given equation of line with y = mx + c, we get ⇒ m 1 = - 1/3 supreme fashion s.a